maths.free › Calculus › 1. Understanding the Derivative › The tangent line approximation

The tangent line approximation

Among all functions, linear functions are simplest. One of the powerful consequences of a function y = f(x) being differentiable at a point (a,f(a)) is that, up close, the function y = f(x) is locally linear and looks…

Introduction

Among all functions, linear functions are simplest. One of the powerful consequences of a function \(y = f(x)\) being differentiable at a point \((a,f(a))\) is that, up close, the function \(y = f(x)\) is locally linear and looks like its tangent line at that point. In certain circumstances, this allows us to approximate the original function \(f\) with a simpler function \(L\) that is linear: this can be advantageous when we have limited information about \(f\) or when \(f\) is computationally or algebraically complicated. We will explore all of these situations in what follows.

It is essential to recall that when \(f\) is differentiable at \(x = a\), the value of \(f'(a)\) provides the slope of the tangent line to \(y = f(x)\) at the point \((a,f(a))\). If we know both a point on the line and the slope of the line we can find the equation of the tangent line and write the equation in point-slope form Recall that a line with slope \(m\) that passes through \((x_0,y_0)\) has equation \(y - y_0 = m(x - x_0)\), and this is the point-slope form of the equation. .

Exploration

The tangent line

Given a function \(f\) that is differentiable at \(x = a\), we know that we can determine the slope of the tangent line to \(y = f(x)\) at \((a,f(a))\) by computing \(f'(a)\). The equation of the resulting tangent line is given in point-slope form by \[\begin{aligned}\end{aligned}\]. Note well: there is a major difference between \(f(a)\) and \(f(x)\) in this context. The former is a constant that results from using the given fixed value of \(a\), while the latter is the general expression for the rule that defines the function. The same is true for \(f'(a)\) and \(f'(x)\): we must carefully distinguish between these expressions. Each time we find the tangent line, we need to evaluate the function and its derivative at a fixed \(a\)-value.

In Figure, we see the graph of a function \(f\) and its tangent line at the point \((a,f(a))\). Notice how when we zoom in we see the local linearity of \(f\) more clearly highlighted. The function and its tangent line are nearly indistinguishable up close. Local linearity can also be seen dynamically in this interactive graphic.

The local linearization

A slight change in perspective and notation will enable us to be more precise in discussing how the tangent line approximates \(f\) near \(x = a\). By solving for \(y\), we can write the equation for the tangent line as \[\begin{aligned}\end{aligned}\] This line is itself a function of \(x\). Replacing the variable \(y\) with the expression \(L(x)\), we call \[\begin{aligned}\end{aligned}\] the local linearization of \(f\) at the point \((a,f(a))\). In this notation, \(L(x)\) is nothing more than a new name for the tangent line. As we saw above, for \(x\) close to \(a\), \(f(x) \approx L(x)\).

Example

Suppose that a function \(y = f(x)\) has its tangent line approximation given by \(L(x) = 3 - 2(x-1)\) at the point \((1,3)\), but we do not know anything else about the function \(f\). How can we estimate the value of \(f(x)\) for \(x\) near \(1\)?

Solution

To estimate a value of \(f(x)\) for \(x\) near 1, such as \(f(1.2)\), we can use the fact that \(f(1.2) \approx L(1.2)\) and hence \[\begin{aligned}\end{aligned}\].

We emphasize that \(y = L(x)\) is simply a new name for the tangent line function. Using this new notation and our observation that \(L(x) \approx f(x)\) for \(x\) near \(a\), it follows that we can write \[\begin{aligned}\end{aligned}\].

From Activity, we see that the local linearization \(y = L(x)\) is a linear function that shares two important values with the function \(y = f(x)\) that it is derived from. In particular,

\(L(x) = f(a) + f'(a)(x-a)\)\(L(a) = f(a)\)
\(L\)

Hence, \(L'(x) = f'(a)\) for every value of \(x\), and specifically \(L'(a) = f'(a)\). Therefore, we see that \(L\) is a linear function that has both the same value and the same slope as the function \(f\) at the point \((a,f(a))\).

Thus, if we know the linear approximation \(y = L(x)\) for a function, we know the original function's value and its slope at the point of tangency. What remains unknown, however, is the shape of the function \(f\) at the point of tangency. There are essentially four possibilities, as shown in Figure.

Another important question one that we study in depth in second semester calculus regards how accurate the tangent line approximation remains as we move away from the point of tangency. We explore some related ideas in the next activity.

Condensed — the full section is in Boelkins, Active Calculus.

Summary

The tangent line to a differentiable function \(y = f(x)\) at the point \((a,f(a))\) is given in point-slope form by the equation \[\begin{aligned}\end{aligned}\].
The principle of local linearity tells us that if we zoom in on a point where a function \(y = f(x)\) is differentiable, the function will be indistinguishable from its tangent line. That is, a differentiable function looks linear when viewed up close. We rename the tangent line to be the function \(y = L(x)\), where \(L(x) = f(a) + f'(a)(x-a)\). Thus, \(f(x) \approx L(x)\) for all \(x\) near \(x = a\).
If we know the tangent line approximation \(L(x) = f(a) + f'(a)(x-a)\) to a function \(y=f(x)\), then because \(L(a) = f(a)\) and \(L'(a) = f'(a)\), we also know the values of both the function and its derivative at the point where \(x = a\). In other words, the linear approximation tells us the height and slope of the original function. If, in addition, we know the value of \(f''(a)\), we then know whether the tangent line lies above or below the graph of \(y = f(x)\), depending on the concavity of \(f\).

Practice (9)

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

In what follows, we find the equation of the line through the points \((3,2)\) and \((4,-1)\), and remind ourselves why point-slope form works.
1. Compute the slope \(m\) of the line through the points \((3,2)\) and \((4,-1)\) using the slope formula \(\displaystyle m = \dfrac{y_2 - y_1}{x_2 - x_1}\).
2. A generic point on the same line has coordinates \((x,y)\). Use the slope formula \(\displaystyle m = \dfrac{y_2 - y_1}{x_2 - x_1}\) to compute the slope of the line through the point \((3,2)\) and the generic point \((x,y)\).
3. Why are the slopes you found in parts (a) and (b) equal to each other?
4. Use the result of part (c) to solve for \(y-2\); don't simplify further. What do you notice?
A certain function \(y=p(x)\) has its local linearization at \(a = 3\) given by \(L(x) = -2x + 5\).
1. What are the values of \(p(3)\) and \(p'(3)\)? Why?
2. Estimate the value of \(p(2.79)\).
3. Suppose that \(p''(3) = 0\) and you know that \(p''(x) \lt 0\) for \(x \lt 3\). Is your estimate in (b) too large or too small?
4. Suppose that \(p''(x) \gt 0\) for \(x \gt 3\). Use this fact and the additional information above to sketch an accurate graph of \(y = p(x)\) near \(x = 3\). Include a sketch of \(y = L(x)\) in your work.
Die Antwort aufzeigen
1. Since the value and slope of the tangent line match the value and slope of the function at the point of tangency, we know \(p(3) = L(3) = -1\) and \(p'(3) = L'(3) = -2\).
2. Using the local linearization, \(p(2.79) \approx L(2.79) = -2(2.79) + 5 = -0.58\).
3. Since \(p''(x) \lt 0\) for \(x \lt 3\), we know that \(p\) is concave down for \(x \lt 3\). This means to the left of the point of tangency at \(x = 3\), the tangent line lies above the curve, and thus the estimate in (b) is too large.
4. In the following figure, we see a possible function \(p\) that matches the value and slope of the tangent line at the point \((3,-1)\) and that is concave down to the left of \(x=3\) and concave up to the right.
A potato is placed in an oven, and the potato's temperature \(F\) (in degrees Fahrenheit) at various points in time is taken and recorded in the following table. Time \(t\) is measured in minutes.
\(t\) \(0\) \(10\) \(20\) \(30\) \(40\) \(50\) \(60\)
\(F(t)\) \(65\) \(94.7\) \(141.7\) \(167.6\) \(182.4\) \(197.9\) \(209.3\)
1. Use a central difference to estimate \(F'(30)\). Use this estimate as needed in subsequent questions.
2. Find the local linearization \(y = L(t)\) to the function \(y = F(t)\) at the point where \(a = 30\).
3. Determine an estimate for \(F(33)\) by employing the local linearization.
4. Do you think your estimate in (c) is too large or too small? Why?
Die Antwort aufzeigen
1. The central difference approximation with \(h=10\) gives \[\begin{aligned}\end{aligned}\] degrees per minute.
2. The local linearization at \(a=30\) is \(L(t) = F'(30)(t-30)+F(30) = L(t) \approx 2.065(t-30) + 167.6\).
3. Using the linearization from (b), \(F(33) \approx L(33) \approx 2.065(3) + 167.6 = 173.795\) degrees F.
4. Since the values of \(F(t)\) given in the table appear to be increasing at a decreasing rate near \(t = 30\), it follows that the graph of \(F\) is concave down near \(t = 30\). This implies that the graph of \(L(t)\) (the tangent line to \(F\) at \(t = 30)\) lies above the graph of \(F\). Therefore, the approximation given in part (c) is probably an overestimate.
An object moving along a straight line path has a differentiable position function \(y = s(t)\); \(s(t)\) measures the object's position relative to the origin at time \(t\). It is known that at time \(t = 9\) seconds, the object's position is \(s(9) = 4\) feet (i.e., 4 feet to the right of the origin). Furthermore, the object's instantaneous velocity at \(t = 9\) is \(-1.2\) feet per second, and its acceleration at the same instant is \(0.08\) feet per second per second.
1. Use local linearity to estimate the position of the object at \(t = 9.34\).
2. Is your estimate likely too large or too small? Why?
3. In everyday language, describe the behavior of the moving object at \(t = 9\). Is it moving toward the origin or away from it? Is its velocity increasing or decreasing?
Die Antwort aufzeigen
1. The linearization \(L\) of \(s\) at \(t=a\) is \(L(t) = s(a) + s'(a)(t-a)\). With the information we have we can find the linearization of \(s\) at \(t=9\), using the facts that \(s(9) = 4\) and \(s'(9) = -1.2\). Thus, the linearization of \(s\) at \(t=9\) is \(L(t) = 4 - 1.2(t-9)\). One use of the linearization is that for \(t\) close to \(9\), \(L(t)\) can provide a good approximation to \(s(t)\). So \[\begin{aligned}\end{aligned}\].
2. Since \(s''(4) = 0.08\) is positive, we know that the graph of \(s\) is concave up at \(t=4\). This means that the linearization of \(s\) at \(t=4\) lies below the graph of \(s\) and so \(L(9.34)\) is an underestimate to \(s(9.34)\).
3. Given that \(s'(4)\) is negative, the object is moving back toward its original position. The fact that \(s''(4)\) is positive means that the velocity of the object is increasing at \(t=4\). This means that the object is actually slowing down as it moves toward toward its starting position at \(t=4\).
For a certain function \(f\), its derivative is known to be \(f'(x) = (x-1)e^{-x^2}\). Note that you do not know a formula for \(y = f(x)\).
1. At what \(x\)-value(s) is \(f'(x) = 0\)? Justify your answer algebraically, but include a graph of \(f'\) to support your conclusion.
2. Reasoning graphically, for what intervals of \(x\)-values is \(f''(x) \gt 0\)? What does this tell you about the behavior of the original function \(f\)? Explain.
3. Assuming that \(f(2) = -3\), estimate the value of \(f(1.88)\) by finding and using the tangent line approximation to \(f\) at \(x=2\). Is your estimate larger or smaller than the true value of \(f(1.88)\)? Justify your answer.
Die Antwort aufzeigen
1. Since \(e\) raised to any power is always positive (and thus never \(0\)), it follows that the only value that makes \(f'(x) = (x-1)e^{-x^2} = 0\) is \(x=1\). This is confirmed by plotting \(f'\) where we only see the graph cross the \(x\)-axis at \(x=1\) (though we do see the graph approach the \(x\)-axis) as \(x\) increases or decreases without bound).
2. We know that \(f''(x) \gt 0\) wherever \(f'\) is increasing. From the graph of \(f'\), we see that \(f'\) is increasing approximately on the interval \(-0.37 \lt x \lt 1.37\), and thus \(f\) is concave up on this interval.
3. The tangent line approximation at \(a = 2\) is \(L(x) = f(2) + f'(2)(x-2) \approx -3 + 0.01832(x-2)\). Thus, \(f(1.88) \approx L(1.88) \approx -3 + 0.01832(1.88-2) = -3.00022\). Since \(f''(x) \lt 0\) at and around \(x = 2\), \(f\) is concave down there, and therefore the tangent line lies above the curve. This makes our estimate larger than the true value of \(f(1.88)\).
A certain function \(y=p(x)\) has its local linearization at \(a = 3\) given by \(L(x) = -2x + 5\).
1. What are the values of \(p(3)\) and \(p'(3)\)? Why?
2. Estimate the value of \(p(2.79)\).
3. Suppose that \(p''(3) = 0\) and you know that \(p''(x) \lt 0\) for \(x \lt 3\). Is your estimate in (b) too large or too small?
4. Suppose that \(p''(x) \gt 0\) for \(x \gt 3\). Use this fact and the additional information above to sketch an accurate graph of \(y = p(x)\) near \(x = 3\). Include a sketch of \(y = L(x)\) in your work.
Die Antwort aufzeigen
1. Since the value and slope of the tangent line match the value and slope of the function at the point of tangency, we know \(p(3) = L(3) = -1\) and \(p'(3) = L'(3) = -2\).
2. Using the local linearization, \(p(2.79) \approx L(2.79) = -2(2.79) + 5 = -0.58\).
3. Since \(p''(x) \lt 0\) for \(x \lt 3\), we know that \(p\) is concave down for \(x \lt 3\). This means to the left of the point of tangency at \(x = 3\), the tangent line lies above the curve, and thus the estimate in (b) is too large.
4. In the following figure, we see a possible function \(p\) that matches the value and slope of the tangent line at the point \((3,-1)\) and that is concave down to the left of \(x=3\) and concave up to the right.
A potato is placed in an oven, and the potato's temperature \(F\) (in degrees Fahrenheit) at various points in time is taken and recorded in the following table. Time \(t\) is measured in minutes.
\(t\) \(0\) \(10\) \(20\) \(30\) \(40\) \(50\) \(60\)
\(F(t)\) \(65\) \(94.7\) \(141.7\) \(167.6\) \(182.4\) \(197.9\) \(209.3\)
1. Use a central difference to estimate \(F'(30)\). Use this estimate as needed in subsequent questions.
2. Find the local linearization \(y = L(t)\) to the function \(y = F(t)\) at the point where \(a = 30\).
3. Determine an estimate for \(F(33)\) by employing the local linearization.
4. Do you think your estimate in (c) is too large or too small? Why?
Die Antwort aufzeigen
1. The central difference approximation with \(h=10\) gives \[\begin{aligned}\end{aligned}\] degrees per minute.
2. The local linearization at \(a=30\) is \(L(t) = F'(30)(t-30)+F(30) = L(t) \approx 2.065(t-30) + 167.6\).
3. Using the linearization from (b), \(F(33) \approx L(33) \approx 2.065(3) + 167.6 = 173.795\) degrees F.
4. Since the values of \(F(t)\) given in the table appear to be increasing at a decreasing rate near \(t = 30\), it follows that the graph of \(F\) is concave down near \(t = 30\). This implies that the graph of \(L(t)\) (the tangent line to \(F\) at \(t = 30)\) lies above the graph of \(F\). Therefore, the approximation given in part (c) is probably an overestimate.
An object moving along a straight line path has a differentiable position function \(y = s(t)\); \(s(t)\) measures the object's position relative to the origin at time \(t\). It is known that at time \(t = 9\) seconds, the object's position is \(s(9) = 4\) feet (i.e., 4 feet to the right of the origin). Furthermore, the object's instantaneous velocity at \(t = 9\) is \(-1.2\) feet per second, and its acceleration at the same instant is \(0.08\) feet per second per second.
1. Use local linearity to estimate the position of the object at \(t = 9.34\).
2. Is your estimate likely too large or too small? Why?
3. In everyday language, describe the behavior of the moving object at \(t = 9\). Is it moving toward the origin or away from it? Is its velocity increasing or decreasing?
Die Antwort aufzeigen
1. The linearization \(L\) of \(s\) at \(t=a\) is \(L(t) = s(a) + s'(a)(t-a)\). With the information we have we can find the linearization of \(s\) at \(t=9\), using the facts that \(s(9) = 4\) and \(s'(9) = -1.2\). Thus, the linearization of \(s\) at \(t=9\) is \(L(t) = 4 - 1.2(t-9)\). One use of the linearization is that for \(t\) close to \(9\), \(L(t)\) can provide a good approximation to \(s(t)\). So \[\begin{aligned}\end{aligned}\].
2. Since \(s''(4) = 0.08\) is positive, we know that the graph of \(s\) is concave up at \(t=4\). This means that the linearization of \(s\) at \(t=4\) lies below the graph of \(s\) and so \(L(9.34)\) is an underestimate to \(s(9.34)\).
3. Given that \(s'(4)\) is negative, the object is moving back toward its original position. The fact that \(s''(4)\) is positive means that the velocity of the object is increasing at \(t=4\). This means that the object is actually slowing down as it moves toward toward its starting position at \(t=4\).
For a certain function \(f\), its derivative is known to be \(f'(x) = (x-1)e^{-x^2}\). Note that you do not know a formula for \(y = f(x)\).
1. At what \(x\)-value(s) is \(f'(x) = 0\)? Justify your answer algebraically, but include a graph of \(f'\) to support your conclusion.
2. Reasoning graphically, for what intervals of \(x\)-values is \(f''(x) \gt 0\)? What does this tell you about the behavior of the original function \(f\)? Explain.
3. Assuming that \(f(2) = -3\), estimate the value of \(f(1.88)\) by finding and using the tangent line approximation to \(f\) at \(x=2\). Is your estimate larger or smaller than the true value of \(f(1.88)\)? Justify your answer.
Die Antwort aufzeigen
1. Since \(e\) raised to any power is always positive (and thus never \(0\)), it follows that the only value that makes \(f'(x) = (x-1)e^{-x^2} = 0\) is \(x=1\). This is confirmed by plotting \(f'\) where we only see the graph cross the \(x\)-axis at \(x=1\) (though we do see the graph approach the \(x\)-axis) as \(x\) increases or decreases without bound).
2. We know that \(f''(x) \gt 0\) wherever \(f'\) is increasing. From the graph of \(f'\), we see that \(f'\) is increasing approximately on the interval \(-0.37 \lt x \lt 1.37\), and thus \(f\) is concave up on this interval.
3. The tangent line approximation at \(a = 2\) is \(L(x) = f(2) + f'(2)(x-2) \approx -3 + 0.01832(x-2)\). Thus, \(f(1.88) \approx L(1.88) \approx -3 + 0.01832(1.88-2) = -3.00022\). Since \(f''(x) \lt 0\) at and around \(x = 2\), \(f\) is concave down there, and therefore the tangent line lies above the curve. This makes our estimate larger than the true value of \(f(1.88)\).

Symbols used here

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

imaginary unit
i² = −1.

\approx

approximately equal
Equal to the precision shown, not exactly.

\pi

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

\infty

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

\sin,\ \cos,\ \tan

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

\log_b x,\ \ln x

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

\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).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: The tangent line approximation

What is the formula for the general tangent line approximation to a differentiable function y = f(x) at the point (a,f(a))?
What is the principle of local linearity and what is the local linearization of a differentiable function f at a point (a,f(a))?
How does knowing just the tangent line approximation tell us information about the behavior of the original function itself near the point of approximation? How does knowing the second derivative's value at this point provide us additional knowledge of the original function's behavior?

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.

Versuch es selbst.

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

Mehr in 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

\(t\)	\(0\)	\(10\)	\(20\)	\(30\)	\(40\)	\(50\)	\(60\)
\(F(t)\)	\(65\)	\(94.7\)	\(141.7\)	\(167.6\)	\(182.4\)	\(197.9\)	\(209.3\)