3.5 Mean Value Theorem

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We apply the Mean Value Theorem.

The Mean Value Theorem

The Mean Value Theorem is one of the most far-reaching theorems in calculus. It states that for a continuous and differentiable function, the average rate of change over an interval is attained as an instantaneous rate of change at some point inside the interval. The precise mathematical statement is as follows.

Mean Value Theorem If $f(x)$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ , then the Mean Value Equation: $f'(c) = \frac {f(b) - f(a)}{b-a}$ holds for some value $c$ between $a$ and $b$ .

The left hand side of the Mean Value Theorem equation represents the instantaneous rate of change of $f(x)$ at $x = c$ and the right hand side represents the average rate of change of $f(x)$ over the interval from $x=a$ to $x=b$ .

The Mean Value Equation is also frequently presented in the form: $f(b) - f(a)=f'(c)(b-a).$

As a real life example, if a car averages 57.4 miles per hour on a long trip, then by the MVT the car must have been traveling at exactly 57.4 miles per hour at some instant during the trip.

The MVT is considered an existence theorem because it asserts that there exists a special value $c$ inside the interval $(a,b)$ that satisfies Mean Value Equation $f'(c) = \frac {f(b) - f(a)}{b-a}.$ In our examples, we will determine the exact value of $c$ .

We will verify that the function $f(x) = x^2$ satisfies the hypotheses of the MVT on the interval $[0,3]$ and we will find the special value of $c$ that satisfies the Mean Value Equation. Since $x^2$ is a polynomial function, it is continuous and differentiable on any interval. Hence, it is continuous on the closed interval $[0, 3]$ and differentiable on the open interval $(0, 3)$ . Then, by the MVT, the Mean Value Equation holds for some number $c$ in the interval $(0, 3)$ . To find $c$ , we first compute $\frac {f(b) - f(a)}{b-a},$ then we compute $f'(x)$ and finally we solve the equation $f'(x) = \frac {f(b) - f(a)}{b-a}$ for $x$ , discarding any solutions that do not lie in the interval $(0, 3)$ . Keep in mind that the MVT guarantees that we will find at least one solution in this interval. Let’s begin: $\frac {f(b) - f(a)}{b-a} = \frac {f(3) - f(0)}{3-0}$ $= \frac {3^2 - 0^2}{3}= \frac {9 - 0}{3} = 3.$ Next, $f'(x) = 2x$ and finally, we solve the equation $2x = 3$ which gives $x = \frac {3}{2}.$ Note that the value $\frac 32$ is in the interval $(0,3)$ and so the special value, $c$ , guaranteed to exist by the MVT, is $\frac 32$ in this example.

In the above figure, the blue line in the secant line for $f(x) = x^2$ on the interval $[0, 3]$ , and the red line is the tangent line at $x = \frac 32$ . The Mean Value Equation asserts that these lines are parallel, and this is clear in the figure.

Given that the function $f(x) = x^2 + 2$ satisfies the hypotheses of the MVT on the interval $[1,3]$ , find the value of $c$ in the open interval $(1,3)$ which satisfies the conclusion of the theorem.

Compute $f'(x)$ and $\dfrac {f(3) - f(1)}{3-1}$

Solve $f'(x) = \dfrac {f(3) - f(1)}{3-1}$

The value of $c$ is: $\answer [given]{2}$

Given that the function $f(x) = x^2 -3x + 5$ satisfies the hypotheses of the MVT on the interval $[-1,2]$ , find the value of $c$ in the open interval $(-1,2)$ which satisfies the conclusion of the theorem.

Compute $f'(x)$ and $\dfrac {f(2) - f(-1)}{2-(-1)}$

Solve $f'(x) = \dfrac {f(2) - f(-1)}{2-(-1)}$

The value of $c$ is: $\answer [given]{1/2}$

Given that the function $f(x) = 3x^2 -5x + 8$ satisfies the hypotheses of the MVT on the interval $[1,6]$ , find the value of $c$ in the open interval $(1,6)$ which satisfies the conclusion of the theorem.

Compute $f'(x)$ and $\dfrac {f(6) - f(1)}{6-1}$

Solve $f'(x) = \dfrac {f(6) - f(1)}{6-1}$

The value of $c$ is: $\answer [given]{7/2}$

Given that the function $f(x) = x^3 + 2x -9$ satisfies the hypotheses of the MVT on the interval $[-2,2]$ , find the values of $c$ in the open interval $(-2,2)$ which satisfy the conclusion of the theorem.

Compute $f'(x)$ and $\dfrac {f(2) - f(-2)}{2-(-2)}$

Solve $f'(x) = \dfrac {f(2) - f(-2)}{2-(-2)}$

The values of $c$ in ascending order are: $\answer [given]{-2/\sqrt 3}$ and $\answer [given]{2/\sqrt 3}$

We will verify that the function $f(x) = \sqrt x$ satisfies the hypotheses of the MVT on the interval $[0,4]$ and we will find the special value of $c$ that satisfies the Mean Value Equation. The function $\sqrt x$ is continuous on the interval $[0, \infty )$ and differentiable on the interval $(0, \infty )$ . Hence, it is continuous on the closed interval $[0, 4]$ and differentiable on the open interval $(0, 4)$ . Then, by the MVT, the Mean Value Equation holds for some number $c$ in the interval $(0, 4)$ . To find $c$ , we first compute $\frac {f(b) - f(a)}{b-a} = \frac {\sqrt 4 - \sqrt 0}{4-0} = \frac {2}{4} = \frac {1}{2}.$ Next we compute $f'(x) = \frac {d}{dx} \sqrt x = \frac {d}{dx} x^{1/2} = (1/2)x^{-1/2} = \frac {1}{2\sqrt x}.$ Finally we solve the equation $\frac {1}{2\sqrt x} = \frac {1}{2}$ which gives $\sqrt x = 1$ and so $x=1.$ Note that the value $x = 1$ is in the interval $(0,4)$ and so the special value, $c$ , guaranteed to exist by the MVT, is $1$ in this example.

In the above figure, the blue line in the secant line for $f(x) = \sqrt x$ on the interval $[0, 4]$ , and the red line is the tangent line at $x = 1$ . The Mean Value Equation asserts that these lines are parallel, and this is clear in the figure.

Given that the function $f(x) = \sqrt x$ satisfies the hypotheses of the MVT on the interval $[0,9]$ , find the value of $c$ in the open interval $(0,9)$ which satisfies the conclusion of the theorem.

Compute $f'(x)$ and $\dfrac {f(9) - f(0)}{9-0}$

Solve $f'(x) = \dfrac {f(9) - f(0)}{9-0}$

The value of $c$ is: $\answer [given]{9/4}$

Given that the function $f(x) = \sqrt x$ satisfies the hypotheses of the MVT on the interval $[4,16]$ , find the value of $c$ in the open interval $(4,16)$ which satisfies the conclusion of the theorem.

Compute $f'(x)$ and $\dfrac {f(16) - f(4)}{16-4}$

Solve $f'(x) = \dfrac {f(16) - f(4)}{16-4}$

The value of $c$ is: $\answer [given]{9}$

Given that the function $f(x) = \sqrt {2x+1}$ satisfies the hypotheses of the MVT on the interval $[0,4]$ , find the value of $c$ in the open interval $(0,4)$ which satisfies the conclusion of the theorem.

Compute $f'(x)$ and $\dfrac {f(4) - f(0)}{4-0}$

Solve $f'(x) = \dfrac {f(4) - f(0)}{4-0}$

The value of $c$ is: $\answer [given]{3/2}$

We will verify that the function $f(x) = e^x$ satisfies the hypotheses of the MVT on the interval $[0,2]$ and we will find the special value of $c$ that satisfies the Mean Value Equation. The function $e^x$ is continuous and differentiable on the interval $(-\infty , \infty )$ . Hence, it is continuous on the closed interval $[0, 2]$ and differentiable on the open interval $(0, 2)$ . Then, by the MVT, the Mean Value Equation holds for some number $c$ in the interval $(0, 2)$ . To find $c$ , we first compute $\frac {f(b) - f(a)}{b-a} = \frac {e^2 - e^0}{2-0} = \frac {e^2 - 1}{2}.$ Next we compute $f'(x) = \frac {d}{dx} e^x = e^x.$ Finally we solve the equation $e^x = \frac {e^2 - 1}{2}$ which gives $x = \ln (\frac {e^2 - 1}{2}) \approx 1.16.$

Note that the value $\ln (\frac {e^2 - 1}{2})$ is in the interval $(0,4)$ and so the special value, $c$ , guaranteed to exist by the MVT, is $\ln (\frac {e^2 - 1}{2})$ in this example.

In the above figure, the blue line in the secant line for $f(x) = e^x$ on the interval $[0, 2]$ , and the red line is the tangent line at $x = \ln (\frac {e^2 - 1}{2})$ . The Mean Value Equation asserts that these lines are parallel, and this is clear in the figure.

Given that the function $f(x) = e^{2x}$ satisfies the hypotheses of the MVT on the interval $[0,1]$ , find the value of $c$ in the open interval $(0,1)$ which satisfies the conclusion of the theorem.

Compute $f'(x)$ and $\dfrac {f(1) - f(0)}{1-0}$

Solve $f'(x) = \dfrac {f(1) - f(0)}{1-0}$

The value of $c$ is: $\answer [given]{(1/2) \ln ((e^2 -1)/2)}$

We will verify that the function $f(x) = \ln (x)$ satisfies the hypotheses of the MVT on the interval $[1,e]$ and we will find the special value of $c$ that satisfies the Mean Value Equation. The function $\ln (x)$ is continuous and differentiable on the interval $(0, \infty )$ . Hence, it is continuous on the closed interval $[1, e]$ and differentiable on the open interval $(1, e)$ . Then, by the MVT, the Mean Value Equation holds for some number $c$ in the interval $(1, e)$ . To find $c$ , we first compute $\frac {f(b) - f(a)}{b-a} = \frac {\ln (e) - \ln (1)}{e-1} = \frac {1 - 0}{e-1} = \frac {1}{e-1}.$ Next we compute $f'(x) = \frac {d}{dx} \ln (x) = \frac {1}{x}.$ Finally we solve the equation $\frac {1}{x} = \frac {1}{e-1}$ which gives $x = e-1 \approx 1.72.$

Note that the value $x = e-1$ is in the interval $(1,e)$ and so the special value, $c$ , guaranteed to exist by the MVT, is $e-1$ in this example.

In the above figure, the blue line in the secant line for $f(x) = \ln (x)$ on the interval $[1, e]$ , and the red line is the tangent line at $x = e-1$ . The Mean Value Equation asserts that these lines are parallel, and this is clear in the figure.

Given that the function $f(x) = \ln (x)$ satisfies the hypotheses of the MVT on the interval $[1,e]$ , find the value of $c$ in the open interval $(1,e)$ which satisfies the conclusion of the theorem.

Compute $f'(x)$ and $\dfrac {f(e) - f(1)}{e-1}$

Solve $f'(x) = \dfrac {f(e) - f(1)}{e-1}$

The value of $c$ is: $\answer [given]{e-1}$

Given that the function $f(x) = \sin (x)$ satisfies the hypotheses of the MVT on the interval $[0, \pi ]$ , find the value of $c$ in the open interval $(0, \pi )$ which satisfies the conclusion of the theorem.

Compute $f'(x)$ and $\dfrac {f(\pi ) - f(0)}{\pi - 0}$

Solve $f'(c) = \dfrac {f(\pi ) - f(0)}{\pi - 0}$

The value of $c$ is: $\answer [given]{\pi /2}$

Here is a detailed, lecture style video on the Mean Value Theorem: