Question

Найдите f'(-2) если f(x)= (3-2x)(2x+3)

Answer 1

$f(x)= (3-2x)(2x+3)=f(x)= -(2x-3)(2x+3)=-(4x^2-9)=-4x^2+9\\\\f'(x)=(-4x^2+9)'=-8x\\\\f'(-2)=-8\cdot(-2)=16$

Answer 2

To find the derivative of the function f(x) = (3 - 2x)(2x + 3) and then evaluate it at x = -2, we'll use the product rule and then plug in x = -2.

The product rule states that if you have a function of the form u(x) * v(x), then the derivative (denoted as (u*v)') is given by:

(u*v)' = u' * v + u * v'

where u' represents the derivative of u with respect to x, and v' represents the derivative of v with respect to x.

Let's apply the product rule step by step:

Given f(x) = (3 - 2x)(2x + 3)

Let u(x) = (3 - 2x) and v(x) = (2x + 3).

Now, we find the derivatives of u and v:

u'(x) = d/dx (3 - 2x) = -2 v'(x) = d/dx (2x + 3) = 2

Now, apply the product rule:

f'(x) = u' * v + u * v' f'(x) = (-2) * (2x + 3) + (3 - 2x) * 2

Simplify:

f'(x) = -4x - 6 + 6 - 4x f'(x) = -8x

Now, to find f'(-2), we just substitute x = -2 into the derivative we found:

f'(-2) = -8 * (-2) = 16

So, the value of f'(-2) is 16.