Mean Value Theorem

[Johnmoon]

I need some help completing these problems, my professor isn't any help so I figure this will be better.

 

[Otis]

Hello. Have you read the forum's submission guidelines, yet (i.e., READ BEFORE POSTING)?

We prefer separate threads for separate exercises. Also, it's hard for tutors to guess where a student got stuck, when they can't see any effort.

?

 

[Johnmoon]

I understand it is continuous, I believe the derivative is f'(x)=6x^2-30x-36, I also understand that when you plug in -5 and 10 into the original function, the outcomes aren't equal. How do I find the answer? If I make the function equal to 0, I always end up getting it wrong. please help.

 

[ksdhart2]

when you plug in -5 and 10 into the original function, the outcomes aren't equal.

This bit right here makes me wonder if you're perhaps confusing Rolle's Theorem with the Mean Value Theorem [MVT]. The two theorems are somewhat similar and have nearly identical requirements, but they imply two totally different things. Rolle's Theorem requires that $f(a) = f(b)$, but the MVT does not have this requirement.

The only two requirements for the MVT are that the function be continuous on the closed interval and differentiable on the open interval, both of which you've verified. Given these two criteria, the MVT says:

$\displaystyle f^{\prime}(c) = \frac{f(b) - f(a)}{b - a} \implies f^{\prime}(c)(b - a) = f(b) - f(a)$

Plugging in all of the known values will result in a quadratic in $c$, giving two solutions. However, neither of the solutions you've posted are correct, and the actual answers will be irrational. Without seeing the work you did to arrive at these answers, I can provide no further guidance. Please share with us all of the work you've done on this problem, even the parts you know for sure are wrong. Thank you.

 

[Otis]

… How do I find the answer? …

I see that you've marked this thread 'solved'. Can you post your work and the solution? Future readers may find it useful. Thanks!

?

 

[Johnmoon]

upon plugging -5 and 10 into the function 2x^3-15x^2-36x+2 I got ---- 443 for b(10) and 142 for a(-5). I then do b-a and get 15, and divide 585 by 15. This gets me 39. After simplifying f'(x) to 6(x+1)(x-6) I make them equal to 39 and get 45 and 38. However this happens to still be incorrect.

 

[ksdhart2]

Okay, so if I'm understanding you right, you've correctly identified the values of $a, \: b, \: f(a), \: f(b)$, and $f^{\prime}(c)$, but I'm at a total loss as to how you found your final answers. You've now found two different answers than in your original post, but I can't seem to figure out how you got any of them. Starting from $f^{\prime}(c) = 39$, we have:

$\displaystyle 6(c + 1)(c - 6) = 39$

In order to solve for $c$ in this particular instance, it would probably be helpful to expand the left hand side, so let's do that now:

$\displaystyle 6c^2 - 30c - 36 = 39$

To solve a quadratic equation, we want the right-hand side to be 0. Let's subtract 39 from both sides:

$\displaystyle 6c^2 - 30c - 75 = 0$

At this point, it seems only logical to try to use the Quadratic Formula, which tells us that:

$\displaystyle c = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

What do you think the appropriate values of $A, \: B, \:$ and $C$ would be, based on the quadratic equation we found? What happens if you plug those values in and simplify things down?

 

[Otis]

? $\;$ Here's a tip: We can reduce the coefficients in the quadratic equation, before applying the Quadratic Formula, because each is a multiple of 3:

6c² - 30c - 75 = 0

(3)(2)c² - (3)(10)c - (3)(25) = 0

Divide each side by 3.

2c² - 10c - 25 = 0

The solutions will be the same, either way. Using the smaller coefficients makes simplifying the answer a wee bit easier.

$\;$