MVT: f(x) = -2x^4+2x^2+3x on [-2,2]?

Question

Find two values (c) lying on (-2,2) that satisfy the conclusion of MVT (i.e. x = 0)

My answer:
f'(x)= -8x^3+7x
f(-2) = -30
f(2) = -18

so, -18 - (-30) = -8x^3+7x(2 - (-2))
-48 = -32x^3+28x

But I only get one zero and I think I messed up in the equation somewhere. Please advise :)

Oh ya forgot f' c(x) -> x = 1 * c

Thanks.

Answer 1

Where did you get f '(x) = - 8x^3 + 7x?

Shouldn't it be f '(x) = - 8x^3 + 4x + 3?

BUT the MVT says f '(c) = [ f(b) - f(a)]/ (b - a).

So you should start again.

Answer 2

One zero is at (0,0). The other is at (sqrt(7/8),0).