>>84124
>thoroughly
Not entirely sure what you're asking, but I'll take a crack at it anyway.

f(x) = 2x^2 + 8x^2 + 2x - 11 = 10x^2 + 2x - 11
so...
f'(x) = 20x + 2.

f'(x) = 4x only at the x where:

20x + 2 = 4x
16x = -2
x = -(1/8)

>>84124
>f'(x)=4x
>f'(x)=20x+2
>f'(x) does not equal f'(x)

Is your teacher high?

>>84124
f(x) = 2x^2 + 8x^2 + 2x - 11
f(x) = 10x^2 + 2x - 11
Integrate so increase power by 1 then divide by the new power
f'(x) = 10/3x^3 + x^2 - 11x
4x = 10/3x^3 + x^2 - 11x
10/3x^3 + x^2 - 15x = 0
Differentiate it to get
3* 10/3x^2 + 2*x - 15 = 0
10x^2 + 2x - 15 = 0
(x + 5)(x - 3) = 0

Therefore x = -5 and x = 3

>>84172
>integrate
>f'
What are you doing?

>>84176
Oh fuck I'm retarded it's been a few years

Anyway looking back at the question, did OP mean f(x) = 2x^3 + 8x^2 + 2x - 11? Don't see why it would be asked as 2x^2 + 8x^2

>>84124
What kind of exercise is that?

>>84124
f'(x) = 4x when f(x) = 2x^2 + 8x^2 + 2x - 11

2(4x)(4x) + 8(4x)(4x) + 2(4x) - 11

32x^2 + 128x^2 + 8x - 11

160x^2 + 8x - 11 <-- This is your answer