Question

find the derivative of the function f(x) =2x²+3x-5 at x=-1.Also prove that f'(0)+3f'(-1)=0

Answer 1

f(-1)=2(-1)^2+3(-1)-5
=2(1)-3-5
=2-8
=-6

f(0)'+3f'(-1)=0
0-3f'=0
f'=0/-3
f'=0

Answer 2

Answer:

$f'(-1) = -1.$

Step-by-step explanation:

Given:

$f(x)=2x^2+3x-5$.

The derivative of this function is given by

$f'(x) = \dfrac{\mathrm{d}f(x)}{\mathrm{d} x}\\=\dfrac{\mathrm{d}}{\mathrm{d} x}\left(2x^2+3x-5 \right)\\=2\cdot 2x+3\\=4x+3.$

At x = -1,

$f'(x)= 4\cdot (-1)+3=-4+3=-1.$.

To prove:

$f'(0)+3f'(-1)=0$.

We have,

$f'(0) = 4\cdot 0+3 = 3.\\f'(-1) = 4\cdot (-1) +3=-1\\\therefore f'(0) +3f'(-1) = 3+3(-1) = 3-3=0.$

Thus, LHS = RHS, proved.