# Using the Mean Value Theorem?

If $f(x)=5x^2+3x-10$ ... Find the average slope of the function $f$ on the interval $[-1,1]$ and ... Verify the MVT by finding a number $c$ in $(-1,1)$ such that $f'(c)=m$

I thought I understood the MVT but I can't seem to solve any problems. This is a good example. Can someone walk me through it so I can see where I went wrong?

Allow me to clarify: I evaluated $f(x)$ for -1 and 1, and got -8 and -2, respectively. I substituted those into the $f(b)-f(a)/b-a$ formula. Should I have evaluated $f'(x)$? What do I do?

Another edit: Can anyone simply walk me through the problem? At this point, I need the most basic instructions possible. I need to see someone solve for f(1) and f(-1), apply those values to the MVT formula, and then solve to find c.

• It would be good to see what you've tried to point out where you went wrong :) – Yiyuan Lee Oct 20 '14 at 19:53
• @YiyuanLee I evaluated f(x) for -1 and 1, getting -8 and -2, respectively. I tried to substitute these values into the $(f(b)-f(a))/(b-a)$ formula, and it came out to 3, which I'm being told is incorrect. Since I can't get past the first part, I haven't even tried the second. – Adam R Oct 20 '14 at 19:56

## 2 Answers

Taking derivative of $f(x)$, you should have the slope on the interval, which is:

$$f'(x)=10x+3$$

In order to calculate the average slope, just take $\frac {f'(1)+f'(-1)}{2}$

Remember the Mean Value Theorem: Suppose f is continuous on [a,b] and if $f(a)$ $\lt$ $m$ $\lt$ $f(b)$, then there exists $c \in [a,b]$ such that $f(c)=m$.

You shall see that since $f'(x)=10x+3$ is continuous on $[-1,1]$ and $f'(-1)=-10+3=-7$$\lt$ $m$ $\lt$ $f'(1)=10+3=13$, by Mean Value Theorem, you can find any $c$ $\in$ $[-1,1]$ such that $f'(c)=m$ in between $-7$ and $13$.

Answer to your edit I: You should take $f'$ and apply Mean Value Theorem since it will give you a better picture of using it rather than being too absorbed in using the formula given.

• Oh, I see... thank you! – Adam R Oct 20 '14 at 20:28

The average slope is $\frac{f(1) - f(-1)}{2}$ then $f^\prime (x) = 10x + 3$ so simply solve the equation $10c + 3 = \frac{f(1) - f(-1)}{2}$ for $c$.