# Thread: Mean Value Theorem

1. ## Mean Value Theorem

I have to use the Mean Value Theorem to find f'(c)
1. x^3+x on [1,2] the answer says c=1.5275...
2. x^4+2 on[-1,2] the answer says c=1.0772...

The entire directions state: Verify that the given function f satisfies the hypothesis of the MVT on the given interval [a,b]. Then find all numbers cbetween a and b for which: (f(b)-f(a)/b-a)=f'(c)

For 1, I found the deriv. which is 3x^2+1
Then applied 1 and 2 to the original equation and got 4 and 13 respectively.
plugged them into the MVT equation:13-4/2-1=9.
Then I plugged the 9 with the derivitive 3x^2+1 for it to become 3x^2+1-9=3x^2-8.
Made x stand by itself and got x = +or-(sqrt 8/3). and it doesn't add up to the answer.

For 2, Long story short, I did the exact same steps as in #1 and the result was the same.

2. Originally Posted by driver327
I have to use the Mean Value Theorem to find f'(c)
1. x^3+x on [1,2] the answer says c=1.52775...
2. x^4+2 on[-1,2] the answer says c=1.0772...

For 1, I found the deriv. which is 3x^2+1 , applied 1 and 2 to the original equation and got 4 and 13 respectively. plugged them into the MVT equation:
13-4/2-1=9. I plugged the 9 with the derivitive 3x^2+1 for it to become 3x^2+1-9=3x^2-8. make x stand by itself and got x= +or-(sqrt 8/3). and it doesn't add up to the answer.

For 2, Long story short, I did the exact same steps as in #1 and the result was the same.
This is very badly expressed, what you seem to be finding is the point in the
given interval where the derivative is equal to the slope of the line joining
(x1,f(x1)) to (x2,f(x2)), where the given interval is (x1,x2).

(by the way for the first c=1.5275...)

RonL

3. You're almost there. You just made a slight error.

$f(1)=2, \;\ f(2)=10, \;\ f'(x)=3x^{2}+1, \;\ f'(c)=3c^{2}+1$

But, $f'(c)=\frac{f(b)-f(a)}{b-a}=8$

$3c^{2}+1=8$

$c=\pm\sqrt{\frac{7}{3}}=\pm\frac{\sqrt{21}}{3}$

Only one of these is in the interval.

4. ## one more, screwed up on quotient rule or something...

1/x+1 on [0,2]. the answer says .73...
The deriv. is -1/(x+1)^2 with the quotient rule. f(0)=1 and f(2)=1/3 then MVT , the answer came as -2/3/2 or -2/6=-1/3.
I plugged that with the derivitive: -1/(x+1)^2+1/3.
I made them like to get rid of the denominators to get -3+x^2+1.
x^2=2
x=+or-(sqrt 2)
And I know I did not get this right again.
Oh yeah, disregard 2. I got it.

5. $f(0)=1, \;\ f(2)=\frac{1}{3}, \;\ f'(c)=\frac{-1}{(c+1)^{2}}, \;\$

$f'(c)=\frac{\frac{1}{3}-1}{2-0}=\frac{-1}{3}$

$\frac{-1}{(c+1)^{2}}=\frac{-1}{3}$

Solve for c. Just do it the same way I showed in the other post.

6. I got all of that. The problem is getting the c to be alone since it is (x+1)^2 I am working with. I have attempted quadratic but it did not work.
I have -1/(c+1)^2=-1/3
mult. the (c+1)^2 to be -1=-1/3(c+1)^2
I broke the binomial for the right side to be (-1/3)c^2-(2/3)c+1.
put 1 on other side: -2=(-1/3)c^2-(2/3)c
-2=(c-1)(-1/3)c
I know what your trying to say though, it is the (c+1)^2 that is in the way.

7. Oh, you're having trouble with solving the equation. I see.

I don't see anything in the way.

$\frac{-1}{(c+1)^{2}}=\frac{-1}{3}$

$-1=\frac{-1}{3}(c+1)^{2}$

$3=(c+1)^{2}$

$c^{2}+2c-2=0$

$c=-(\sqrt{3}+1), \;\ \sqrt{3}-1$