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Inverse Functions

Verify inverse functions.

Inverse Functions

  1. Find and evaluate composite functions (IA 10.1.1).
  2. Determine whether a function is one-to-one (IA 10.1.2).

A composite function is a two-step function and can have numerical or variable inputs.

\(\left(fog\right)\left(x\right)=f\left(g\left(x\right)\right)\) is read as “f of g of x”.

To evaluate a composite function, we always start by evaluating the inner function and then evaluate the outer function in terms of the inner function.

Find and evaluate composite functions.

Try it.

For functions \(f\left(x\right)=2x-7\) , \(g\left(x\right)=\frac{x+7}{2}\) , find:

  1. ⓐ \(g\left(5\right)\)
  2. ⓑ \(f\left(g\left(5\right)\right)\)
  3. ⓒ \(f\left(g\left(x\right)\right)\)
Solution
  1. ⓐ To find \(g\left(5\right)\) , we evaluate \(g\left(x\right)\) when \(x\) is 5.
    \(g\left(x\right)=\frac{x+7}{2}\)
    \(g\left(5\right)=\frac{5+7}{2}=\frac{12}{2}=6\)
  2. ⓑTo find \(f\left(g\left(5\right)\right)\) , we start evaluating the inner function \(g\) in terms of 5 (see part a) and then evaluate the outer function \(f\) in terms of this value.
    \(f(g\left(5\right))=f(6)=2(6)-7=12-7=5\)
  3. ⓒIn parts a and b we had numerical outputs because our inputs were numbers. When we find \(f\left(g\left(x\right)\right)\) this will be a function written in terms of the variable x.
    \(f(g\left(x\right))=f(\frac{x+7}{2})=2(\frac{x+7}{2})-7=x+7-7=x\)
    This is interesting, notice that the functions \(f(x)\) and \(g(x)\) have a special relationship in that one undoes the other. We call functions like this, inverses of one another. For any one-to-one function \(f(x)\) , the inverse is a function \({f}^{-1}(x)\) such that \({f}^{-1}(f(x))=x\) .

Find and evaluate composite functions.

For each of the following function pairs find:

Try it.

\(f\left(x\right)=\sqrt[3]{x-2}\) , \(g\left(x\right)={x}^{3}+2\)

ⓐ \(f(g\left(x\right))\)

ⓑ \(g(f\left(x\right))\)

ⓒ Graph the functions f(x) and g(x) on the same coordinate system below

ⓓ What do you notice about the relationship between the graphs of f(x) and g(x)?

Try it.

\(f(x)=\frac{1}{(x+3)}\) , \(g(x)=\frac{1}{x}-3\)

ⓐ \(f(g\left(x\right))\)

ⓑ \(g(f\left(x\right))\)

ⓒ Graph the functions f(x) and g(x) on the same coordinate system below

ⓓ What do you notice about the relationship between the graphs of f(x) and g(x)?

Condensed — the full section is in OpenStax College Algebra 2e.

Verifying That Two Functions Are Inverse Functions

Betty is traveling to Milan for a fashion show and wants to know what the temperature will be. She is not familiar with the Celsius scale. To get an idea of how temperature measurements are related, Betty wants to convert 75 degrees Fahrenheit to degrees Celsius using the formula

\[C=\frac{5}{9}(F-32)\]

and substitutes 75 for \(F\) to calculate

\[\frac{5}{9}(75-32)\approx 24\text{^{\circ}C}\]

Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, Betty gets the week’s weather forecast from for Milan, and wants to convert all of the temperatures to degrees Fahrenheit.

At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for \(F\) after substituting a value for \(C.\) For example, to convert 26 degrees Celsius, she could write

\[\begin{array}{lll}26 & = & \frac{5}{9}(F-32) \\ 26⋅\frac{9}{5} & = & F-32 \\ F & = & 26⋅\frac{9}{5}+32\approx 79\end{array}\]

After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature.

The formula for which Betty is searching corresponds to the idea of an inverse function, which is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function.

Given a function \(f(x),\) we represent its inverse as \({f}^{-1}(x),\) read as \(“f\) inverse of \(x.\text{”}\) The raised \(-1\) is part of the notation. It is not an exponent; it does not imply a power of \(-1\) . In other words, \({f}^{-1}(x)\) does not mean \(\frac{1}{f(x)}\) because \(\frac{1}{f(x)}\) is the reciprocal of \(f\) and not the inverse.

\[({f}^{-1}∘f)(x)={f}^{-1}(f(x))={f}^{-1}(y)=x\]\[({f}^{-1}∘f)(x)={f}^{-1}(4x)=\frac{1}{4}(4x)=x\]\[({f}^{}∘{f}^{-1})(x)=f(\frac{1}{4}x)=4(\frac{1}{4}x)=x\]
Example

Try it.

If for a particular one-to-one function \(f(2)=4\) and \(f(5)=12,\) what are the corresponding input and output values for the inverse function?

Solution

The inverse function reverses the input and output quantities, so if

\[\begin{array}{lll}f(2) & = & 4,\ \text{then }{f}^{-1}(4)=2; \\ f(5) & = & 12,{\ \text{then f}}^{-1}(12)=5.\end{array}\]

Alternatively, if we want to name the inverse function \(g,\) then \(g(4)=2\) and \(g(12)=5.\)

Condensed — the full section is in OpenStax College Algebra 2e.

Finding Domain and Range of Inverse Functions

The outputs of the function \(f\) are the inputs to \({f}^{-1},\) so the range of \(f\) is also the domain of \({f}^{-1}.\) Likewise, because the inputs to \(f\) are the outputs of \({f}^{-1},\) the domain of \(f\) is the range of \({f}^{-1}.\) We can visualize the situation as in .

When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of \(f(x)=\sqrt{x}\) is \({f}^{-1}(x)={x}^{2},\) because a square “undoes” a square root; but the square is only the inverse of the square root on the domain \([0,\infty ),\) since that is the range of \(f(x)=\sqrt{x}.\)

We can look at this problem from the other side, starting with the square (toolkit quadratic) function \(f(x)={x}^{2}.\) If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and –3. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the “inverse” is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.

In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function \(f(x)={x}^{2}\) with its domain limited to \([0,\infty ),\) which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).

If \(f(x)={(x-1)}^{2}\) on \([1,\infty ),\) then the inverse function is \({f}^{-1}(x)=\sqrt{x}+1.\)

  • The domain of \(f\) = range of \({f}^{-1}\) = \([1,\infty ).\)
  • The domain of \({f}^{-1}\) = range of \(f\) = \([0,\infty ).\)

Condensed — the full section is in OpenStax College Algebra 2e.

Finding and Evaluating Inverse Functions

Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases.

Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range.

Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function.

Example

Try it.

A function \(f(t)\) is given in , showing distance in miles that a car has traveled in \(t\) minutes. Find and interpret \({f}^{-1}(70).\)

\(t\ \text{(minutes)}\) 30507090
\(f(t)\ \text{(miles)}\) 20406070
Solution

The inverse function takes an output of \(f\) and returns an input for \(f.\) So in the expression \({f}^{-1}(70),\) 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function \(f,\) 90 minutes, so \({f}^{-1}(70)=90.\) The interpretation of this is that, to drive 70 miles, it took 90 minutes.

Alternatively, recall that the definition of the inverse was that if \(f(a)=b,\) then \({f}^{-1}(b)=a.\) By this definition, if we are given \({f}^{-1}(70)=a,\) then we are looking for a value \(a\) so that \(f(a)=70.\) In this case, we are looking for a \(t\) so that \(f(t)=70,\) which is when \(t=90.\)

Condensed — the full section is in OpenStax College Algebra 2e.

Finding Inverse Functions and Their Graphs

Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function \(f(x)={x}^{2}\) restricted to the domain \([0,\infty )\text{,}\) on which this function is one-to-one, and graph it as in .

Restricting the domain to \([0,\infty )\) makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.

We already know that the inverse of the toolkit quadratic function is the square root function, that is, \({f}^{-1}(x)=\sqrt{x}.\) What happens if we graph both \(f\) and \({f}^{-1}\) on the same set of axes, using the \(x\text{-}\) axis for the input to both \(f\ \text{and }{f}^{-1}?\)

We notice a distinct relationship: The graph of \({f}^{-1}(x)\) is the graph of \(f(x)\) reflected about the diagonal line \(y=x,\) which we will call the identity line, shown in .

This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.

Example

Try it.

Given the graph of \(f(x)\) in , sketch a graph of \({f}^{-1}(x).\)

Solution

This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of \((0,\infty )\) and range of \((-\infty ,\infty ),\) so the inverse will have a domain of \((-\infty ,\infty )\) and range of \((0,\infty ).\)

If we reflect this graph over the line \(y=x,\) the point \((1,0)\) reflects to \((0,1)\) and the point \((4,2)\) reflects to \((2,4).\) Sketching the inverse on the same axes as the original graph gives .

Key Concepts

  • If \(g(x)\) is the inverse of \(f(x),\) then \(g(f(x))=f(g(x))=x.\) See , , and .
  • Only some of the toolkit functions have an inverse. See .
  • For a function to have an inverse, it must be one-to-one (pass the horizontal line test).
  • A function that is not one-to-one over its entire domain may be one-to-one on part of its domain.
  • For a tabular function, exchange the input and output rows to obtain the inverse. See .
  • The inverse of a function can be determined at specific points on its graph. See .
  • To find the inverse of a formula, solve the equation \(y=f(x)\) for \(x\) as a function of \(y.\) Then exchange the labels \(x\) and \(y.\) See , , and .
  • The graph of an inverse function is the reflection of the graph of the original function across the line \(y=x.\) See .

Practice (40)

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

  1. For functions \(f\left(x\right)=2x-7\) , \(g\left(x\right)=\frac{x+7}{2}\) , find:

    1. ⓐ \(g\left(5\right)\)
    2. ⓑ \(f\left(g\left(5\right)\right)\)
    3. ⓒ \(f\left(g\left(x\right)\right)\)
    Die Antwort aufzeigen
    1. ⓐ To find \(g\left(5\right)\) , we evaluate \(g\left(x\right)\) when \(x\) is 5.
      \(g\left(x\right)=\frac{x+7}{2}\)
      \(g\left(5\right)=\frac{5+7}{2}=\frac{12}{2}=6\)
    2. ⓑTo find \(f\left(g\left(5\right)\right)\) , we start evaluating the inner function \(g\) in terms of 5 (see part a) and then evaluate the outer function \(f\) in terms of this value.
      \(f(g\left(5\right))=f(6)=2(6)-7=12-7=5\)
    3. ⓒIn parts a and b we had numerical outputs because our inputs were numbers. When we find \(f\left(g\left(x\right)\right)\) this will be a function written in terms of the variable x.
      \(f(g\left(x\right))=f(\frac{x+7}{2})=2(\frac{x+7}{2})-7=x+7-7=x\)
      This is interesting, notice that the functions \(f(x)\) and \(g(x)\) have a special relationship in that one undoes the other. We call functions like this, inverses of one another. For any one-to-one function \(f(x)\) , the inverse is a function \({f}^{-1}(x)\) such that \({f}^{-1}(f(x))=x\) .

  2. \(f\left(x\right)=\sqrt[3]{x-2}\) , \(g\left(x\right)={x}^{3}+2\)

    ⓐ \(f(g\left(x\right))\)

    ⓑ \(g(f\left(x\right))\)

    ⓒ Graph the functions f(x) and g(x) on the same coordinate system below

    ⓓ What do you notice about the relationship between the graphs of f(x) and g(x)?

  3. \(f(x)=\frac{1}{(x+3)}\) , \(g(x)=\frac{1}{x}-3\)

    ⓐ \(f(g\left(x\right))\)

    ⓑ \(g(f\left(x\right))\)

    ⓒ Graph the functions f(x) and g(x) on the same coordinate system below

    ⓓ What do you notice about the relationship between the graphs of f(x) and g(x)?

  4. Determine ⓐ whether each graph is the graph of a function and, if so, ⓑ whether it is one-to-one.

    Die Antwort aufzeigen

    ⓐ Since any vertical line intersects the graph in at most one point, the graph is the graph of a function. Since any horizontal line intersects the graph in at most one point, the graph is the graph of a one-to-one function.

    ⓑ Since any vertical line intersects the graph in at most one point, the graph is the graph of a function. However, a horizontal line shown on the graph may intersect it in two points. This graph does not represent a one-to-one function.

  5. If for a particular one-to-one function \(f(2)=4\) and \(f(5)=12,\) what are the corresponding input and output values for the inverse function?

    Die Antwort aufzeigen

    The inverse function reverses the input and output quantities, so if

    \[\begin{array}{lll}f(2) & = & 4,\ \text{then }{f}^{-1}(4)=2; \\ f(5) & = & 12,{\ \text{then f}}^{-1}(12)=5.\end{array}\]

    Alternatively, if we want to name the inverse function \(g,\) then \(g(4)=2\) and \(g(12)=5.\)

  6. Given that \({h}^{-1}(6)=2,\) what are the corresponding input and output values of the original function \(h?\)

    Die Antwort aufzeigen

    \(h(2)=6\)

  7. If \(f(x)=\frac{1}{x+2}\) and \(g(x)=\frac{1}{x}-2,\) is \(g={f}^{-1}?\)

    Die Antwort aufzeigen
    \[\begin{array}{lll}g(f(x)) & = & \frac{1}{(\frac{1}{x+2})}-2 \\ & = & x+2-2 \\ & = & x\end{array}\]

    We must also verify the other formula.

    \[\begin{array}{lll}f(g(x)) & = & \frac{1}{\frac{1}{x}-2+2} \\ & = & \frac{1}{\frac{1}{x}} \\ & = & x\end{array}\]

    so

    \[g={f}^{-1}\ \text{and }f={g}^{-1}\]

  8. If \(f(x)={x}^{3}-4\) and \(g(x)=\sqrt[\ 3]{x+4},\) is \(g={f}^{-1}?\)

    Die Antwort aufzeigen

    Yes

  9. If \(f(x)={x}^{3}\) (the cube function) and \(g(x)=\frac{1}{3}x,\) is \(g={f}^{-1}?\)

    Die Antwort aufzeigen
    \[f(g(x))=\frac{{x}^{3}}{27}\ne x\]

    No, the functions are not inverses.

  10. If \(f(x)={(x-1)}^{3}\ \text{and}\ g(x)=\sqrt[3]{x}+1,\) is \(g={f}^{-1}?\)

    Die Antwort aufzeigen

    Yes

  11. Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed in . We restrict the domain in such a fashion that the function assumes all y-values exactly once.

    ConstantIdentityQuadraticCubicReciprocal
    \(f(x)=c\) \(f(x)=x\) \(f(x)={x}^{2}\) \(f(x)={x}^{3}\) \(f(x)=\frac{1}{x}\)
    Reciprocal squaredCube rootSquare rootAbsolute value
    \(f(x)=\frac{1}{{x}^{2}}\) \(f(x)=\sqrt[3]{x}\) \(f(x)=\sqrt{x}\) \(f(x)=|x|\)
    Die Antwort aufzeigen

    The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no inverse.

    The absolute value function can be restricted to the domain \([0,\infty ),\) where it is equal to the identity function.

    The reciprocal-squared function can be restricted to the domain \((0,\infty ).\)

  12. The domain of function \(f\) is \((1,\infty )\) and the range of function \(f\) is \((-\infty ,-2).\) Find the domain and range of the inverse function.

    Die Antwort aufzeigen

    The domain of function \({f}^{-1}\) is \((-\infty \text{,}-2)\) and the range of function \({f}^{-1}\) is \((1,\infty ).\)

  13. A function \(f(t)\) is given in , showing distance in miles that a car has traveled in \(t\) minutes. Find and interpret \({f}^{-1}(70).\)

    \(t\ \text{(minutes)}\) 30507090
    \(f(t)\ \text{(miles)}\) 20406070
    Die Antwort aufzeigen

    The inverse function takes an output of \(f\) and returns an input for \(f.\) So in the expression \({f}^{-1}(70),\) 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function \(f,\) 90 minutes, so \({f}^{-1}(70)=90.\) The interpretation of this is that, to drive 70 miles, it took 90 minutes.

    Alternatively, recall that the definition of the inverse was that if \(f(a)=b,\) then \({f}^{-1}(b)=a.\) By this definition, if we are given \({f}^{-1}(70)=a,\) then we are looking for a value \(a\) so that \(f(a)=70.\) In this case, we are looking for a \(t\) so that \(f(t)=70,\) which is when \(t=90.\)

  14. Using , find and interpret ⓐ \(\ f(60),\) and ⓑ \(\ {f}^{-1}(60).\)

    \(t\ \text{(minutes)}\) 3050607090
    \(f(t)\ \text{(miles)}\) 2040506070
    Die Antwort aufzeigen
    1. ⓐ \(f(60)=50.\) In 60 minutes, 50 miles are traveled.
    2. ⓑ \({f}^{-1}(60)=70.\) To travel 60 miles, it will take 70 minutes.

  15. A function \(g(x)\) is given in . Find \(g(3)\) and \({g}^{-1}(3).\)

    Die Antwort aufzeigen

    To evaluate \(g(3),\) we find 3 on the x-axis and find the corresponding output value on the y-axis. The point \((3,1)\) tells us that \(g(3)=1.\)

    To evaluate \({g}^{-1}(3),\) recall that by definition \({g}^{-1}(3)\) means the value of x for which \(g(x)=3.\) By looking for the output value 3 on the vertical axis, we find the point \((5,3)\) on the graph, which means \(g(5)=3,\) so by definition, \({g}^{-1}(3)=5.\) See .

  16. Using the graph in , ⓐ find \({g}^{-1}(1),\) and ⓑ estimate \({g}^{-1}(4).\)

    Die Antwort aufzeigen
    1. ⓐ 3
    2. ⓑ 5.6

  17. Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.

    \[C=\frac{5}{9}(F-32)\]
    Die Antwort aufzeigen
    \[\begin{array}{lll}C & = & \frac{5}{9}(F-32) \\ C⋅\frac{9}{5} & = & F-32 \\ F & = & \frac{9}{5}C+32\end{array}\]

    By solving in general, we have uncovered the inverse function. If

    \[C=h(F)=\frac{5}{9}(F-32),\]

    then

    \[F={h}^{-1}(C)=\frac{9}{5}C+32\]

    In this case, we introduced a function \(h\) to represent the conversion because the input and output variables are descriptive, and writing \({C}^{-1}\) could get confusing.

  18. Solve for \(x\) in terms of \(y\) given \(y=\frac{1}{3}(x-5).\)

    Die Antwort aufzeigen

    \(x=3y+5\)

  19. Find the inverse of the function \(f(x)=\frac{2}{x-3}+4.\)

    Die Antwort aufzeigen
    \[\begin{array}{llll}y & = & \frac{2}{x-3}+4 & \ \text{Set up an equation}. \\ y-4 & = & \frac{2}{x-3} & \ \text{Subtract 4 from both sides}. \\ x-3 & = & \frac{2}{y-4} & \ \text{Multiply both sides by }x-3\ \text{and divide by }y-4. \\ x & = & \frac{2}{y-4}+3 & \ \text{Add 3 to both sides}.\end{array}\]

    So \({f}^{-1}(y)=\frac{2}{y-4}+3\) or \({f}^{-1}(x)=\frac{2}{x-4}+3.\)

  20. Find the inverse of the function \(f(x)=2+\sqrt{x-4}.\)

    Die Antwort aufzeigen
    \[\begin{array}{lll}y & = & 2+\sqrt{x-4} \\ {(y-2)}^{2} & = & x-4 \\ x & = & {(y-2)}^{2}+4\end{array}\]

    So \({f}^{-1}(x)={(x-2)}^{2}+4.\)

    The domain of \(f\) is \([4,\infty ).\) Notice that the range of \(f\) is \([2,\infty ),\) so this means that the domain of the inverse function \({f}^{-1}\) is also \([2,\infty ).\)

  21. What is the inverse of the function \(f(x)=2-\sqrt{x}?\) State the domains of both the function and the inverse function.

    Die Antwort aufzeigen

    \({f}^{-1}(x)={(2-x)}^{2};\) \(\text{domain}\ \text{of}\ f:\ [0,\infty );\) \(\text{domain}\ \text{of}\ {f}^{-1}:\ (-\infty ,2]\)

  22. Given the graph of \(f(x)\) in , sketch a graph of \({f}^{-1}(x).\)

    Die Antwort aufzeigen

    This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of \((0,\infty )\) and range of \((-\infty ,\infty ),\) so the inverse will have a domain of \((-\infty ,\infty )\) and range of \((0,\infty ).\)

    If we reflect this graph over the line \(y=x,\) the point \((1,0)\) reflects to \((0,1)\) and the point \((4,2)\) reflects to \((2,4).\) Sketching the inverse on the same axes as the original graph gives .

  23. Draw graphs of the functions \(f\) and \(\ {f}^{-1}\) from .

  24. Describe why the horizontal line test is an effective way to determine whether a function is one-to-one?

    Die Antwort aufzeigen

    Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that \(y\) -values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no \(y\) -values repeat and the function is one-to-one.

  25. Why do we restrict the domain of the function \(f(x)={x}^{2}\) to find the function’s inverse?

  26. Can a function be its own inverse? Explain.

    Die Antwort aufzeigen

    Yes. For example, \(f(x)=\frac{1}{x}\) is its own inverse.

  27. Are one-to-one functions either always increasing or always decreasing? Why or why not?

  28. How do you find the inverse of a function algebraically?

    Die Antwort aufzeigen

    Given a function \(y=f(x),\) solve for \(x\) in terms of \(y.\) Interchange the \(x\) and \(y.\) Solve the new equation for \(y.\) The expression for \(y\) is the inverse, \(y={f}^{-1}(x).\)

  29. Show that the function \(f(x)=a-x\) is its own inverse for all real numbers \(a.\)

  30. \(f(x)=x+3\)

    Die Antwort aufzeigen

    \({f}^{-1}(x)=x-3\)

  31. \(f(x)=x+5\)

  32. \(f(x)=2-x\)

    Die Antwort aufzeigen

    \({f}^{-1}(x)=2-x\)

  33. \(f(x)=3-x\)

  34. \(f(x)=\frac{x}{x+2}\)

    Die Antwort aufzeigen

    \({f}^{-1}(x)=\frac{-2x}{x-1}\)

  35. \(f(x)=\frac{2x+3}{5x+4}\)

  36. \(f(x)={(x+7)}^{2}\)

    Die Antwort aufzeigen

    domain of \(f(x):\ [-7,\infty );\ {f}^{-1}(x)=\sqrt{x}-7\)

  37. \(f(x)={(x-6)}^{2}\)

  38. \(f(x)={x}^{2}-5\)

    Die Antwort aufzeigen

    domain of \(f(x):\ [0,\infty );\ {f}^{-1}(x)=\sqrt{x+5}\)

  39. Given \(f(x)=\frac{x}{2+x}\) and \(g(x)=\frac{2x}{1-x}:\)

    1. ⓐFind \(f(g(x))\) and \(g(f(x)).\)
    2. ⓑWhat does the answer tell us about the relationship between \(f(x)\) and \(g(x)?\)
    Die Antwort aufzeigen
    1. ⓐ\(f(g(x))=x\) and \(g(f(x))=x.\)
    2. ⓑThis tells us that \(f\) and \(g\) are inverse functions

  40. \(f(x)=\sqrt[3]{x-1}\) and \(g(x)={x}^{3}+1\)

    Die Antwort aufzeigen

    \(f(g(x))=x,\ g(f(x))=x\)

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.
^\circ
degrees
1/360 of a full turn. 180° = π radians.
\approx
approximately equal
Equal to the precision shown, not exactly.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
\sup,\ \inf
supremum, infimum
Least upper bound, greatest lower bound.
\pm
plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.
\neq
not equal
The two sides are different.
|x|
absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
i
imaginary unit
i² = −1.
\log_b x,\ \ln x
logarithm, natural log
The exponent b must be raised to for x; ln uses base e.
\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}
number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Inverse Functions

  1. Verify inverse functions.
  2. Determine the domain and range of an inverse function, and restrict the domain of a function to make it one-to-one.
  3. Find or evaluate the inverse of a function.
  4. Use the graph of a one-to-one function to graph its inverse function on the same axes.
  5. Find and evaluate composite functions (IA 10.1.1).
  6. Determine whether a function is one-to-one (IA 10.1.2).

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Versuch es selbst.

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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