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Basic Classes of Functions

Calculate the slope of a linear function and interpret its meaning.

Linear Functions and Slope

The easiest type of function to consider is a linear function. Linear functions have the form $f(x)=ax+b,$ where $a$ and $b$ are constants. In , we see examples of linear functions when $a$ is positive, negative, and zero. Note that if $a>0,$ the graph of the line rises as $x$ increases. In other words, $f(x)=ax+b$ is increasing on $\text{(-\infty , \infty )}.$ If $a<0,$ the graph of the line falls as $x$ increases. In this case, $f(x)=ax+b$ is decreasing on $\text{(-\infty , \infty )}.$ If $a=0,$ the line is horizontal.

As suggested by , the graph of any linear function is a line. One of the distinguishing features of a line is its slope. The slope is the change in $y$ for each unit change in $x.$ The slope measures both the steepness and the direction of a line. If the slope is positive, the line points upward when moving from left to right. If the slope is negative, the line points downward when moving from left to right. If the slope is zero, the line is horizontal. To calculate the slope of a line, we need to determine the ratio of the change in $y$ versus the change in $x.$ To do so, we choose any two points $({x}_{1},{y}_{1})$ and $({x}_{2},{y}_{2})$ on the line and calculate $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}.$ In , we see this ratio is independent of the points chosen.

We now examine the relationship between slope and the formula for a linear function. Consider the linear function given by the formula $f(x)=ax+b.$ As discussed earlier, we know the graph of a linear function is given by a line. We can use our definition of slope to calculate the slope of this line. As shown, we can determine the slope by calculating $({y}_{2}-{y}_{1})\text{/}({x}_{2}-{x}_{1})$ for any points $({x}_{1},{y}_{1})$ and $({x}_{2},{y}_{2})$ on the line. Evaluating the function $f$ at $x=0,$ we see that $(0,b)$ is a point on this line. Evaluating this function at $x=1,$ we see that $(1,a+b)$ is also a point on this line. Therefore, the slope of this line is

\[\frac{(a+b)-b}{1-0}=a.\]

We have shown that the coefficient $a$ is the slope of the line. We can conclude that the formula $f(x)=ax+b$ describes a line with slope $a.$ Furthermore, because this line intersects the $y$-axis at the point $(0,b),$ we see that the $y$-intercept for this linear function is $(0,b).$ We conclude that the formula $f(x)=ax+b$ tells us the slope, $a,$ and the $y$-intercept, $(0,b),$ for this line. Since we often use the symbol $m$ to denote the slope of a line, we can write

\[f(x)=mx+b\]\[f(x)-{y}_{1}=m(x-{x}_{1}).\]

Condensed — the full section is in OpenStax Calculus Volume 1.

Polynomials

A linear function is a special type of a more general class of functions: polynomials. A polynomial function is any function that can be written in the form

\[f(x)={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\text{\ldots }+{a}_{1}x+{a}_{0}\]

for some integer $n\ge 0$ and constants ${a}_{n},{a}_{n-1}\text{,\ldots ,}{a}_{0},$ where ${a}_{n}\ne 0.$ In the case when $n=0,$ we allow for ${a}_{0}=0;$ if ${a}_{0}=0,$ the function $f(x)=0$ is called the zero function. The value $n$ is called the degree of the polynomial; the constant ${a}_{n}$ is called the leading coefficient. A linear function of the form $f(x)=mx+b$ is a polynomial of degree 1 if $m\ne 0$ and degree 0 if $m=0.$ A polynomial of degree 0 is also called a constant function. A polynomial function of degree 2 is called a quadratic function. In particular, a quadratic function has the form $f(x)=a{x}^{2}+bx+c,$ where $a\ne 0.$ A polynomial function of degree $3$ is called a cubic function.

Some polynomial functions are power functions. A power function is any function of the form $f(x)=a{x}^{b},$ where $a$ and $b$ are any real numbers. The exponent in a power function can be any real number, but here we consider the case when the exponent is a positive integer. (We consider other cases later.) If the exponent is a positive integer, then $f(x)=a{x}^{n}$ is a polynomial. If $n$ is even, then $f(x)=a{x}^{n}$ is an even function because $f(\text{-}\text{x})=a{(\text{-}\text{x})}^{n}=a{x}^{n}$ if $n$ is even. If $n$ is odd, then $f(x)=a{x}^{n}$ is an odd function because $f(\text{-}\text{x})=a{(\text{-}\text{x})}^{n}=\text{-}a{x}^{n}$ if $n$ is odd ().

Condensed — the full section is in OpenStax Calculus Volume 1.

Algebraic Functions

By allowing for quotients and fractional powers in polynomial functions, we create a larger class of functions. An algebraic function is one that involves addition, subtraction, multiplication, division, rational powers, and roots. Two types of algebraic functions are rational functions and root functions.

Just as rational numbers are quotients of integers, rational functions are quotients of polynomials. In particular, a rational function is any function of the form $f(x)=p(x)\text{/}q(x),$ where $p(x)$ and $q(x)$ are polynomials. For example,

\[f(x)=\frac{3x-1}{5x+2}\ \text{and}\ g(x)=\frac{4}{{x}^{2}+1}\]

are rational functions. A root function is a power function of the form $f(x)={x}^{1\text{/}n},$ where $n$ is a positive integer greater than one. For example, $f(x)={x}^{1\text{/}2}=\sqrt{x}$ is the square-root function and $g(x)={x}^{1\text{/}3}=\sqrt[3]{x}$ is the cube-root function. By allowing for compositions of root functions and rational functions, we can create other algebraic functions. For example, $f(x)=\sqrt{4-{x}^{2}}$ is an algebraic function.

Example

Try it.

For each of the following functions, find the domain and range.

$f(x)=\frac{3x-1}{5x+2}$
$f(x)=\sqrt{4-{x}^{2}}$

Solution

It is not possible to divide by zero, so the domain is the set of real numbers $x$ such that $x\ne \text{-}2\text{/}5.$ To find the range, we need to find the values $y$ for which there exists a real number $x$ such that
\[y=\frac{3x-1}{5x+2}.\]
When we multiply both sides of this equation by $5x+2,$ we see that $x$ must satisfy the equation
\[5xy+2y=3x-1.\]

From this equation, we can see that $x$ must satisfy
\[2y+1=x(3-5y).\]
If $y=3\text{/}5,$ this equation has no solution. On the other hand, as long as $y\ne 3\text{/}5,$
\[x=\frac{2y+1}{3-5y}\]
satisfies this equation. We can conclude that the range of $f$ is $\{y|y\ne 3\text{/}5\}.$
To find the domain of $f$, we need to identify values of $x$ with $4-{x}^{2}\ge 0$. Note that $4-{x}^{2}=0$ when $x=2$ or $x=-2$. Solving $4-{x}^{2}>0$ is equivalent to solving $(2-x)(2+x)>0$. If this inequality is true, then both factors are positive or both are negative.

Case 1: Both factors are positive.
\[2-x>0\text{and}2+x>0\] \[x<2\text{and}x>-2\]
So, the expression $4-{x}^{2}$ is positive for all values in the interval $-2
Case 2: Both factors are negative.
\[2-x<0\text{and}2+x<0\] \[x>2\text{and}x<-2\]
There is no value of \(x$ that can satisfy both inequalities.

The domain of the function is $\{x|-2\le x\le 2\}$.

To identify the range, note that if $-2\le x\le 2$, then $0\le 4-{x}^{2}\le 4$. Therefore, $0\le \sqrt{4-{x}^{2}}\le 2$. The range of $f$ is $\{y|0\le y\le 2\}$.

Condensed — the full section is in OpenStax Calculus Volume 1.

Transcendental Functions

Thus far, we have discussed algebraic functions. Some functions, however, cannot be described by basic algebraic operations. These functions are known as transcendental functions because they are said to “transcend,” or go beyond, algebra. The most common transcendental functions are trigonometric, exponential, and logarithmic functions. A trigonometric function relates the ratios of two sides of a right triangle. They are $\sin x,\cos x,\tan x,\text{cot}x,\text{sec}x,\text{and}\ \text{csc}x.$ (We discuss trigonometric functions later in the chapter.) An exponential function is a function of the form $f(x)={b}^{x},$ where the base $b>0,b\ne 1.$ A logarithmic function is a function of the form $f(x)={\log }_{b}(x)$ for some constant $b>0,b\ne 1,$ where ${\log }_{b}(x)=y$ if and only if ${b}^{y}=x.$ (We also discuss exponential and logarithmic functions later in the chapter.)

Example

Try it.

Classify each of the following functions, a. through c., as algebraic or transcendental.

$f(x)=\frac{\sqrt{{x}^{3}+1}}{4x+2}$
$f(x)={2}^{{x}^{2}}$
$f(x)=\text{sin}(2x)$

Solution

Since this function involves basic algebraic operations only, it is an algebraic function.
This function cannot be written as a formula that involves only basic algebraic operations, so it is transcendental. (Note that algebraic functions can only have powers that are rational numbers.)
As in part b., this function cannot be written using a formula involving basic algebraic operations only; therefore, this function is transcendental.

Piecewise-Defined Functions

Sometimes a function is defined by different formulas on different parts of its domain. A function with this property is known as a piecewise-defined function. The absolute value function is an example of a piecewise-defined function because the formula changes with the sign of $x\text{:}$

\[f(x)=\{\begin{array}{l}\text{-}x,x<0 \\ x,x\ge 0\end{array}.\]

Other piecewise-defined functions may be represented by completely different formulas, depending on the part of the domain in which a point falls. To graph a piecewise-defined function, we graph each part of the function in its respective domain, on the same coordinate system. If the formula for a function is different for $xa,$ we need to pay special attention to what happens at $x=a$ when we graph the function. Sometimes the graph needs to include an open or closed circle to indicate the value of the function at $x=a.$ We examine this in the next example.

Example

Try it.

Sketch a graph of the following piecewise-defined function:

\[f(x)=\{\begin{array}{l}x+3,\ x<1 \\ {(x-2)}^{2},\ x\ge 1\end{array}.\]

Solution

Graph the linear function $y=x+3$ on the interval $(\text{-\infty },1)$ and graph the quadratic function $y={(x-2)}^{2}$ on the interval $[1,\infty ).$ Since the value of the function at $x=1$ is given by the formula $f(x)={(x-2)}^{2},$ we see that $f(1)=1.$ To indicate this on the graph, we draw a closed circle at the point $(1,1).$ The value of the function is given by $f(x)=x+3$ for all $x<1,$ but not at $x=1.$ To indicate this on the graph, we draw an open circle at $(1,4).$

Example

Try it.

In a big city, drivers are charged variable rates for parking in a parking garage. They are charged $10 for the first hour or any part of the first hour and an additional $2 for each hour or part thereof up to a maximum of $30 for the day. The parking garage is open from 6 a.m. to 12 midnight.

Write a piecewise-defined function that describes the cost $C$ to park in the parking garage as a function of hours parked $x.$
Sketch a graph of this function $C(x).$

Solution

Since the parking garage is open 18 hours each day, the domain for this function is \(\text{\{}x|0 \[C(x)=\{\begin{array}{l}10,0
The graph of the function consists of several horizontal line segments.

Transformations of Functions

We have seen several cases in which we have added, subtracted, or multiplied constants to form variations of simple functions. In the previous example, for instance, we subtracted 2 from the argument of the function $y={x}^{2}$ to get the function $f(x)={(x-2)}^{2}.$ This subtraction represents a shift of the function $y={x}^{2}$ two units to the right. A shift, horizontally or vertically, is a type of transformation of a function. Other transformations include horizontal and vertical scalings, and reflections about the axes.

A vertical shift of a function occurs if we add or subtract the same constant to each output $y.$ For $c>0,$ the graph of $f(x)+c$ is a shift of the graph of $f(x)$ up $c$ units, whereas the graph of $f(x)-c$ is a shift of the graph of $f(x)$ down $c$ units. For example, the graph of the function $f(x)={x}^{2}+4$ is the graph of $y={x}^{2}$ shifted up $4$ units; the graph of the function $f(x)={x}^{2}-4$ is the graph of $y={x}^{2}$ shifted down $4$ units ().

A horizontal shift of a function occurs if we add or subtract the same constant to each input $x.$ For $c>0,$ the graph of $f(x+c)$ is a shift of the graph of $f(x)$ to the left $c$ units; the graph of $f(x-c)$ is a shift of the graph of $f(x)$ to the right $c$ units. Why does the graph shift left when adding a constant and shift right when subtracting a constant? To answer this question, let’s look at an example.

Consider the function $f(x)=|x+3|$ and evaluate this function at $x-3.$ Since $f(x-3)=|x|$ and \(x-3

A vertical scaling of a graph occurs if we multiply all outputs $y$ of a function by the same positive constant. For $c>0,$ the graph of the function $cf(x)$ is the graph of $f(x)$ scaled vertically by a factor of $c.$ If $c>1,$ the values of the outputs for the function $cf(x)$ are larger than the values of the outputs for the function $f(x);$ therefore, the graph has been stretched vertically. If \(0

The horizontal scaling of a function occurs if we multiply the inputs $x$ by the same positive constant. For $c>0,$ the graph of the function $f(cx)$ is the graph of $f(x)$ scaled horizontally by a factor of $c.$ If $c>1,$ the graph of $f(cx)$ is the graph of $f(x)$ compressed horizontally. If \(0

Horizontal shift of the graph of $y=f(x).$ If $b>0,$ shift left. If $b<0,$ shift right.
Horizontal scaling of the graph of $y=f(x+b)$ by a factor of $|a|.$ If $a<0,$ reflect the graph about the $y$-axis.
Vertical scaling of the graph of $y=f(a(x+b))$ by a factor of $|c|.$ If $c<0,$ reflect the graph about the $x$-axis.
Vertical shift of the graph of $y=cf(a(x+b)).$ If $d>0,$ shift up. If $d<0,$ shift down.

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

The power function $f(x)={x}^{n}$ is an even function if $n$ is even and $n\ne 0,$ and it is an odd function if $n$ is odd.
The root function $f(x)={x}^{1\text{/}n}$ has the domain $[0,\infty )$ if $n$ is even and the domain $(\text{-\infty },\infty )$ if $n$ is odd. If $n$ is odd, then $f(x)={x}^{1\text{/}n}$ is an odd function.
The domain of the rational function $f(x)=p(x)\text{/}q(x),$ where $p(x)$ and $q(x)$ are polynomial functions, is the set of $x$ such that $q(x)\ne 0.$
Functions that involve the basic operations of addition, subtraction, multiplication, division, and powers are algebraic functions. All other functions are transcendental. Trigonometric, exponential, and logarithmic functions are examples of transcendental functions.
A polynomial function $f$ with degree $n\ge 1$ satisfies $f(x)\to \text{\pm }\infty$ as $x\to \text{\pm }\infty .$ The sign of the output as $x\to \infty$ depends on the sign of the leading coefficient only and on whether $n$ is even or odd.
Vertical and horizontal shifts, vertical and horizontal scalings, and reflections about the $x$- and $y$-axes are examples of transformations of functions.

Key Equations

Point-slope equation of a line	$y-{y}_{1}=m(x-{x}_{1})$
Slope-intercept form of a line	$y=mx+b$
Standard form of a line	$ax+by=c$
Polynomial function	$f(x)={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\text{\cdots }+{a}_{1}x+{a}_{0}$

Basic Classes of Functions

For the following exercises, for each pair of points, a. find the slope of the line passing through the points and b. indicate whether the line is increasing, decreasing, horizontal, or vertical.

For the following exercises, write the equation of the line satisfying the given conditions in slope-intercept form.

For the following exercises, for each linear equation, a. give the slope $m$ and $y$-intercept b, if any, and b. graph the line.

For the following exercises, for each polynomial, a. find the degree; b. find the zeros, if any; c. find the $y$-intercept(s), if any; d. use the leading coefficient to determine the graph’s end behavior; and e. determine algebraically whether the polynomial is even, odd, or neither.

For the following exercises, use the graph of $f(x)={x}^{2}$ to graph each transformed function $g.$

For the following exercises, use the graph of $f(x)=\sqrt{x}$ to graph each transformed function $g.$

For the following exercises, use the graph of $y=f(x)$ to graph each transformed function $g.$

Condensed — the full section is in OpenStax Calculus Volume 1.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Consider the line passing through the points $(11,-4)$ and $(-4,5),$ as shown in .
1. Find the slope of the line.
2. Find an equation for this linear function in point-slope form.
3. Find an equation for this linear function in slope-intercept form.
जवाफ प्रकट गर्नुहोस्
1. The slope of the line is
  \[m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{5-(-4)}{-4-11}=-\frac{9}{15}=-\frac{3}{5}.\]
2. To find an equation for the linear function in point-slope form, use the slope $m=-3\text{/}5$ and choose any point on the line. If we choose the point $(11,-4),$ we get the equation
  \[f(x)+4=-\frac{3}{5}(x-11).\]
3. To find an equation for the linear function in slope-intercept form, solve the equation in part b. for $f(x).$ When we do this, we get the equation
  \[f(x)=-\frac{3}{5}x+\frac{13}{5}.\]
Consider the line passing through points $(-3,2)$ and $(1,4).$ Find the slope of the line.
Find an equation of that line in point-slope form. Find an equation of that line in slope-intercept form.

जवाफ प्रकट गर्नुहोस्

$m=1\text{/}2.$ The point-slope form is
$y-4=\frac{1}{2}(x-1).$
The slope-intercept form is
$y=\frac{1}{2}x+\frac{7}{2}.$
Jessica leaves her house at 5:50 a.m. and goes for a 9-mile run. She returns to her house at 7:08 a.m. Answer the following questions, assuming Jessica runs at a constant pace.
1. Describe the distance $D$ (in miles) Jessica runs as a linear function of her run time $t$ (in minutes).
2. Sketch a graph of $D.$
3. Interpret the meaning of the slope.
जवाफ प्रकट गर्नुहोस्
1. At time $t=0,$ Jessica is at her house, so $D(0)=0.$ At time $t=78$ minutes, Jessica has finished running $9$ mi, so $D(78)=9.$ The slope of the linear function is
  \[m=\frac{9-0}{78-0}=\frac{3}{26}.\]
  The $y$-intercept is $(0,0),$ so the equation for this linear function is
  \[D(t)=\frac{3}{26}t.\]
2. To graph $D,$ use the fact that the graph passes through the origin and has slope $m=3\text{/}26.$
3. The slope $m=3\text{/}26\approx 0.115$ describes the distance (in miles) Jessica runs per minute, or her average velocity.
For the following functions a. and b., i. describe the behavior of $f(x)$ as $x\to \text{\pm }\infty ,$ ii. find all zeros of $f,$ and iii. sketch a graph of $f.$
1. $f(x)=-2{x}^{2}+4x-1$
2. $f(x)={x}^{3}-3{x}^{2}-4x$
जवाफ प्रकट गर्नुहोस्
1. The function $f(x)=-2{x}^{2}+4x-1$ is a quadratic function.
  Because $a=-2<0,\text{as}\ x\to \text{\pm }\infty ,f(x)\to \text{-\infty .}$
  To find the zeros of $f,$ use the quadratic formula. The zeros are
  \[x=\frac{-4\pm \sqrt{{4}^{2}-4(-2)(-1)}}{2(-2)}=\frac{-4\pm \sqrt{8}}{-4}=\frac{-4\pm 2\sqrt{2}}{-4}=\frac{2\pm \sqrt{2}}{2}.\]
  To sketch the graph of $f,$ use the information from your previous answers and combine it with the fact that the graph is a parabola opening downward.
2. The function $f(x)={x}^{3}-3{x}^{2}-4x$ is a cubic function.
  Because $a=1>0,\text{as}\ x\to \infty ,f(x)\to \infty .$ As $x\to \text{-\infty },f(x)\to \text{-\infty }.$
  To find the zeros of $f,$ we need to factor the polynomial. First, when we factor $x$ out of all the terms, we find
  \[f(x)=x({x}^{2}-3x-4).\]
  Then, when we factor the quadratic function ${x}^{2}-3x-4,$ we find
  \[f(x)=x(x-4)(x+1).\]
  Therefore, the zeros of $f$ are $x=0,4,-1.$
  Combining the results from parts i. and ii., draw a rough sketch of $f.$
Consider the quadratic function $f(x)=3{x}^{2}-6x+2.$ Find the zeros of $f.$ Does the parabola open upward or downward?

जवाफ प्रकट गर्नुहोस्

The zeros are $x=1\pm \sqrt{3}\text{/}3.$ The parabola opens upward.
A company is interested in predicting the amount of revenue it will receive depending on the price it charges for a particular item. Using the data from , the company arrives at the following quadratic function to model revenue $R$ (in thousands of dollars) as a function of price per item $p\text{:}$
\[R(p)=p\cdot (-1.04p+26)=-1.04{p}^{2}+26p\]
for $0\le p\le 25.$
1. Predict the revenue if the company sells the item at a price of $p=\text{\$}5$ and $p=\text{\$}17.$
2. Find the zeros of this function and interpret the meaning of the zeros.
3. Sketch a graph of $R.$
4. Use the graph to determine the value of $p$ that maximizes revenue. Find the maximum revenue.
जवाफ प्रकट गर्नुहोस्
1. Evaluating the revenue function at $p=5$ and $p=17,$ we can conclude that
  \[\begin{array}{l} \\ R(5)=-1.04{(5)}^{2}+26(5)=104,\ \text{so revenue}\ =\ \text{\$104,000;} \\ R(17)=-1.04{(17)}^{2}+26(17)=141.44,\ \text{so revenue}\ =\ \text{\$141,440.}\end{array}\]
2. The zeros of this function can be found by solving the equation $-1.04{p}^{2}+26p=0.$ When we factor the quadratic expression, we get $p(-1.04p+26)=0.$ The solutions to this equation are given by $p=0,25.$ For these values of $p,$ the revenue is zero. When $p=\text{\$}0,$ the revenue is zero because the company is giving away its merchandise for free. When $p=\text{\$}25,$ the revenue is zero because the price is too high, and no one will buy any items.
3. Knowing the fact that the function is quadratic, we also know the graph is a parabola. Since the leading coefficient is negative, the parabola opens downward. One property of parabolas is that they are symmetric about the axis, so since the zeros are at $p=0$ and $p=25,$ the parabola must be symmetric about the line halfway between them, or $p=12.5.$
4. The function is a parabola with zeros at $p=0$ and $p=25,$ and it is symmetric about the line $p=12.5,$ so the maximum revenue occurs at a price of $p=\text{\$}12.50$ per item. At that price, the revenue is $R(p)=-1.04{(12.5)}^{2}+26(12.5)=\text{\$}162,500.$
For each of the following functions, find the domain and range.
1. $f(x)=\frac{3x-1}{5x+2}$
2. $f(x)=\sqrt{4-{x}^{2}}$
जवाफ प्रकट गर्नुहोस्
1. It is not possible to divide by zero, so the domain is the set of real numbers $x$ such that $x\ne \text{-}2\text{/}5.$ To find the range, we need to find the values $y$ for which there exists a real number $x$ such that
  \[y=\frac{3x-1}{5x+2}.\]
  When we multiply both sides of this equation by $5x+2,$ we see that $x$ must satisfy the equation
  \[5xy+2y=3x-1.\]
  
  From this equation, we can see that $x$ must satisfy
  \[2y+1=x(3-5y).\]
  If $y=3\text{/}5,$ this equation has no solution. On the other hand, as long as $y\ne 3\text{/}5,$
  \[x=\frac{2y+1}{3-5y}\]
  satisfies this equation. We can conclude that the range of $f$ is $\{y|y\ne 3\text{/}5\}.$
2. To find the domain of $f$, we need to identify values of $x$ with $4-{x}^{2}\ge 0$. Note that $4-{x}^{2}=0$ when $x=2$ or $x=-2$. Solving $4-{x}^{2}>0$ is equivalent to solving $(2-x)(2+x)>0$. If this inequality is true, then both factors are positive or both are negative.
  
  Case 1: Both factors are positive.
  \[2-x>0\text{and}2+x>0\] \[x<2\text{and}x>-2\]
  So, the expression $4-{x}^{2}$ is positive for all values in the interval $-2
  Case 2: Both factors are negative.
  \[2-x<0\text{and}2+x<0\] \[x>2\text{and}x<-2\]
  There is no value of \(x$ that can satisfy both inequalities.
  
  The domain of the function is $\{x|-2\le x\le 2\}$.
  
  To identify the range, note that if $-2\le x\le 2$, then $0\le 4-{x}^{2}\le 4$. Therefore, $0\le \sqrt{4-{x}^{2}}\le 2$. The range of $f$ is $\{y|0\le y\le 2\}$.
Find the domain and range for the function $f(x)=(5x+2)\text{/}(2x-1).$

जवाफ प्रकट गर्नुहोस्

The domain is the set of real numbers $x$ such that $x\ne 1\text{/}2.$ The range is the set $\{y|y\ne 5\text{/}2\}.$
For each of the following functions, determine the domain of the function.
1. $f(x)=\frac{3}{{x}^{2}-1}$
2. $f(x)=\frac{2x+5}{3{x}^{2}+4}$
3. $f(x)=\sqrt{4-3x}$
4. $f(x)=\sqrt[3]{2x-1}$
जवाफ प्रकट गर्नुहोस्
1. You cannot divide by zero, so the domain is the set of values $x$ such that ${x}^{2}-1\ne 0.$ Therefore, the domain is $\{x|x\ne \text{\pm }1\}.$
2. You need to determine the values of $x$ for which the denominator is zero. Since $3{x}^{2}+4\ge 4$ for all real numbers $x,$ the denominator is never zero. Therefore, the domain is $(\text{-\infty },\infty ).$
3. Since the square root of a negative number is not a real number, the domain is the set of values $x$ for which $4-3x\ge 0.$ Therefore, the domain is $\{x|x\le 4\text{/}3\}.$
4. The cube root is defined for all real numbers, so the domain is the interval $\text{(-\infty , \infty ).}$
Find the domain for each of the following functions: $f(x)=(5-2x)\text{/}({x}^{2}+2)$ and $g(x)=\sqrt{5x-1}.$

जवाफ प्रकट गर्नुहोस्

The domain of $f$ is $\text{(-\infty , \infty ).}$ The domain of $g$ is $\{x|x\ge 1\text{/}5\}.$
Classify each of the following functions, a. through c., as algebraic or transcendental.
1. $f(x)=\frac{\sqrt{{x}^{3}+1}}{4x+2}$
2. $f(x)={2}^{{x}^{2}}$
3. $f(x)=\text{sin}(2x)$
जवाफ प्रकट गर्नुहोस्
1. Since this function involves basic algebraic operations only, it is an algebraic function.
2. This function cannot be written as a formula that involves only basic algebraic operations, so it is transcendental. (Note that algebraic functions can only have powers that are rational numbers.)
3. As in part b., this function cannot be written using a formula involving basic algebraic operations only; therefore, this function is transcendental.
Is $f(x)=x\text{/}2$ an algebraic or a transcendental function?

जवाफ प्रकट गर्नुहोस्

Algebraic
Sketch a graph of the following piecewise-defined function:
\[f(x)=\{\begin{array}{l}x+3,\ x<1 \\ {(x-2)}^{2},\ x\ge 1\end{array}.\]

जवाफ प्रकट गर्नुहोस्

Graph the linear function $y=x+3$ on the interval $(\text{-\infty },1)$ and graph the quadratic function $y={(x-2)}^{2}$ on the interval $[1,\infty ).$ Since the value of the function at $x=1$ is given by the formula $f(x)={(x-2)}^{2},$ we see that $f(1)=1.$ To indicate this on the graph, we draw a closed circle at the point $(1,1).$ The value of the function is given by $f(x)=x+3$ for all $x<1,$ but not at $x=1.$ To indicate this on the graph, we draw an open circle at $(1,4).$
Sketch a graph of the function
\[f(x)=\{\begin{array}{l}2-x,x\le 2 \\ x+2,x>2\end{array}.\]

जवाफ प्रकट गर्नुहोस्
In a big city, drivers are charged variable rates for parking in a parking garage. They are charged $10 for the first hour or any part of the first hour and an additional $2 for each hour or part thereof up to a maximum of $30 for the day. The parking garage is open from 6 a.m. to 12 midnight.
1. Write a piecewise-defined function that describes the cost $C$ to park in the parking garage as a function of hours parked $x.$
2. Sketch a graph of this function $C(x).$
जवाफ प्रकट गर्नुहोस्
1. Since the parking garage is open 18 hours each day, the domain for this function is \(\text{\{}x|0 \[C(x)=\{\begin{array}{l}10,0
2. The graph of the function consists of several horizontal line segments.
The cost of mailing a letter is a function of the weight of the letter. Suppose the cost of mailing a letter is $49\text{¢}$ for the first ounce and $21\text{¢}$ for each additional ounce. Write a piecewise-defined function describing the cost $C$ as a function of the weight $x$ for \(0

जवाफ प्रकट गर्नुहोस्

\(C(x)=\{\begin{array}{l}49,0
For each of the following functions, a. and b., sketch a graph by using a sequence of transformations of a well-known function.
1. $f(x)=\text{-}|x+2|-3$
2. $f(x)=3\sqrt{\text{-}\text{x}}+1$
जवाफ प्रकट गर्नुहोस्
1. Starting with the graph of $y=|x|,$ shift $2$ units to the left, reflect about the $x$-axis, and then shift down 3 units.
2. Starting with the graph of $y=\sqrt{x},$ reflect about the $y$-axis, stretch the graph vertically by a factor of 3, and move up 1 unit.
Describe how the function $f(x)=\text{-}{(x+1)}^{2}-4$ can be graphed using the graph of $y={x}^{2}$ and a sequence of transformations.

जवाफ प्रकट गर्नुहोस्

Shift the graph $y={x}^{2}$ to the left 1 unit, reflect about the $x$-axis, then shift down 4 units.
$(-2,4)$ and $(1,1)$

जवाफ प्रकट गर्नुहोस्

a. −1 b. Decreasing
$(-1,4)$ and $(3,-1)$
$(3,5)$ and $(-1,2)$

जवाफ प्रकट गर्नुहोस्

a. 3/4 b. Increasing
$(6,4)$ and $(4,-3)$
$(2,3)$ and $(5,7)$

जवाफ प्रकट गर्नुहोस्

a. 4/3 b. Increasing
$(1,9)$ and $(-8,5)$
$(2,4)$ and $(1,4)$

जवाफ प्रकट गर्नुहोस्

a. 0 b. Horizontal
$(1,4)$ and $(1,0)$
Slope $=-6,$ passes through $(1,3)$

जवाफ प्रकट गर्नुहोस्

$y=-6x+9$
Slope $=3,$ passes through $(-3,2)$
Slope $=\frac{1}{3},$ passes through $(0,4)$

जवाफ प्रकट गर्नुहोस्

$y=\frac{1}{3}x+4$
Slope $=\frac{2}{5},\ x$-intercept $=8$
Passing through $(2,1)$ and $(-2,-1)$

जवाफ प्रकट गर्नुहोस्

$y=\frac{1}{2}x$
Passing through $(-3,7)$ and $(1,2)$
$x$-intercept $=5$ and $y$-intercept $=-3$

जवाफ प्रकट गर्नुहोस्

$y=\frac{3}{5}x-3$
$x$-intercept $=-6$ and $y$-intercept $=9$
$y=-\frac{1}{7}x+1$
$f(x)=-6x$

जवाफ प्रकट गर्नुहोस्

a. $(m=-6,b=0)$ b.
$f(x)=-5x+4$
$6x-5y+15=0$
$f(x)=2{x}^{2}-3x-5$

जवाफ प्रकट गर्नुहोस्

a. 2 b. $\frac{5}{2},-1;$ c. −5 d. As $x\to \pm \infty ,y\to \infty$ e. Neither
$f(x)=-3{x}^{2}+6x$

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

imaginary unit
i² = −1.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

How to: Basic Classes of Functions

Calculate the slope of a linear function and interpret its meaning.
Recognize the degree of a polynomial.
Find the roots of a quadratic polynomial.
Describe the graphs of basic odd and even polynomial functions.
Identify a rational function.
Describe the graphs of power and root functions.
Explain the difference between algebraic and transcendental functions.
Graph a piecewise-defined function.

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

तपाईँको आफ्नै प्रयास गर्नुहोस्

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

यसमा थप Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests

Point-slope equation of a line	\(y-{y}_{1}=m(x-{x}_{1})\)
Slope-intercept form of a line	\(y=mx+b\)
Standard form of a line	\(ax+by=c\)
Polynomial function	\(f(x)={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\text{\cdots }+{a}_{1}x+{a}_{0}\)