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5.1 Using the Mean Value Theorem

3 min readjune 18, 2024

5.1 Using the Mean Value Theorem

In the previous unit, we learned all about applying derivatives to different real-world contexts. What else are derivatives useful for? Turns out, we can also use derivatives to determine and analyze the behaviors of functions! 👀

📈  Mean Value Theorem

The Mean Value Theorem states that if a function f is continuous over the interval [a,b][a, b] and differentiable over the interval (a,b)(a, b), then there exists a point cc within that open interval (a,b)(a,b) where the instantaneous rate of change of the function at cc equals the average rate of change of the function over the interval (a,b)(a, b).

In other words, if a function f is continuous over the interval [a,b][a, b] and differentiable over the interval (a,b)(a, b), there exists some cc on (a,b)(a,b) such that f(c)=f(b)f(a)baf'(c)=\frac{f(b)-f(a)}{b-a}.

Untitled

Image Courtesy of Sumi Vora and Ethan Bilderbeek

Yet another way to phrase this theorem is that if the stated conditions of continuity and differentiability are satisfied, there is a point where the slope of the tangent line is equivalent to the slope of the secant line between aa and bb.

https://www.math.net/img/a/calculus/applications-of-derivatives/mean-value-theorem/mean-value-theorem.png

Image Courtesy of Math.net

Remember from Unit 1, to be continuous over [a,b][a, b] means that there are no holes, asymptotes, or jump discontinuities between points a and b. Because the interval contains closed brackets, the graph must also be continuous at points aa and bb.

Additionally, if we recall from previous guides, to be differentiable over (a,b)(a, b) means that the function is continuous over the interval and that for any point cc over the interval, limxcf(x)f(c)xc\lim_{x\to c} \frac{f(x) - f(c)}{x - c} exists.

✏️ Mean Value Theorem: Walkthrough

We can use the Mean Value Theorem to justify conclusions about functions by applying it over an interval. For example:

Let ff be a differentiable function. The table gives its selected values:

xx33991111
f(x)f(x)202044446767

Can we use the Mean Value Theorem to say the equation f(x)=5f'(x)=5 has a solution where 3<x<93< x<9?

Since it is given that ff is differentiable, we can apply the Mean Value Theorem (MVT) on the interval (3,9)(3,9). This is what we should write out!

Since ff is continuous and differentiable on (3,9)(3,9), MVT can be applied. The MVT states that there exists a cc on (3,9)(3,9) such that f(c)=f(9)f(3)93=442093=246=4.f'(c)=\frac{f(9)-f(3)}{9-3}=\frac{44-20}{9-3}=\frac{24}{6}=4.

454 \neq 5 so MVT cannot be used to say that f(x)=5f'(x)=5 has a solution.


📝 Mean Value Theorem: Practice Problems

Now, it’s time for you to do some practice on your own! 🍀

❓ Mean Value Theorem: Practice

Question 1: Mean Value Theorem

Let h(x)=x3+3x2h(x)=x^3+3x^2 and let cc be the number that satisfies the Mean Value Theorem for hh on the interval [3,0][-3,0].

What is c?c?

Question 2: Mean Value Theorem

Let ff be a differentiable function. The table gives its selected values:

xx227799
f(x)f(x)141443433535

Can we use the Mean Value Theorem to say the equation f(x)=2f'(x)=2 has a solution where 2<x<92<x<9?

✅ Mean Value Theorem: Answers and Solutions

Question 1: Mean Value Theorem

Since hh is a polynomial, hh is continuous on [3,0][-3,0] and differentiable on (3,0)(-3,0). Therefore, the Mean Value Theorem can be applied. By the Mean Value Theorem, there exists a cc on (3,0)(-3,0) such that…

f(c)=f(0)f(3)0(3)=003=0.f'(c)=\frac{f(0)-f(-3)}{0-(-3)}=\frac{0-0}{3}=0.

To find cc, we need to differentiate f(x)f(x) and find cc such that f(c)=0.f'(c)=0.

f(x)=3x2+6xf'(x)=3x^2+6x
3x2+6x=03x^2+6x=0

By the quadratic formula, we have x=2,0x=-2,0.

Since only x=2x=-2 is in the interval (3,0)(-3,0), c=2.c=-2. Great work! Make sure you remember to check if the value(s) you get are in the given interval.

Question 2: Mean Value Theorem

Since it is given that ff is differentiable, we can apply the Mean Value Theorem (MVT) on the interval (2,9)(2,9).

The MVT states that there exists a cc on (2,9)(2,9) such that f(c)=f(9)f(2)92=351492=217=3.f'(c)=\frac{f(9)-f(2)}{9-2}=\frac{35-14}{9-2}=\frac{21}{7}=3. 323 \neq 2 so MVT cannot be used to say that f(x)=2f'(x)=2 has a solution.


💫 Closing

The Mean Value Theorem states that for any continuous function on a closed interval, there exists a value c in the interval such that the value of the derivative of the function at c is equal to the average rate of change of the function over the interval. By using this theorem, we can find the mean value of a function on a given interval, which can provide useful information about the behavior of the function. 🧐