mean value theorem?

Question

I'm pretty sure I understand what's going on but I can't get the right answer for this problem! here's my work

1.) f(x) = x^3 + x^2, [0,1]

d/dx (x^3 + x^2) = 3x^2 + 2x

f(0) = 3(0)^2 + 2(0) = 0
f(1) = 3(1)^2 + 2(1) = 5

f ' (c) = [(5-0) / (1-0)] = 5 so...

3x^2 + 2x - 5 = 0

After using the quadratic equation...I end up with..
(-2 +/- 8) / 6
c = 1 or -1.66

The answer in the book says c = (√ 7 -1) / 3

Is my math wrong?

btw, the point of the problem is to find a value for c

Answer 1

ah, i see what you did.
when you took f(1), you accidently took f'(1).
f(1) = 2
2/1 = 2, the slope of the secant line
so then, do what you did setting the derivative equal to the slope, only this time make it 2.
so you'll get 3x^2 + 2x - 2 = 0
this gives you the answer in the book

Answer 2

C must lie between 0 and 1 so your answer cannot be correct. The problem is that f(0) = 0 and f(1) = 2.
You used the derivative instead of the function itself.
Thus 3x^2 +2x -2 = 0
Then x = ]-2 +/- sqrt(4+4*3*(-2))]/6
= [-2 +/- sqrt(28)]/6
=[-2 +/- 2sqrt(7)]/6
= -1/3 + sqrt(7)/3 which is the book answer.

Answer 3

Problem is f(0)=0 and f(1)=2.

f'(1)=5, but that's not what you want.