Question

1. Given the function f(x) = x3 + 5x2 – X – 4, find...a) f'(-2).b) the values of a such that f'(a) = 56.

Answer 1

Given the function f(x) = x3 + 5x2 – X – 4, find...

a) f'(-2).

b) the values of a such that f'(a) = 56.

step 1

Find f'(x)

first derivative

$f^{\prime}(x)=3x^2+10x-1$

f'(-2)

For x=-2

substitute

$\begin{gathered} f^{\prime}(-2)=3(-2)^2+10(-2)-1 \\ f^{\prime}(-2)=12-20-1 \\ f^{\prime}(-2)=-9 \end{gathered}$

step 2

the values of a such that f'(a) = 56.

$\begin{gathered} f^{\prime}(x)=3x^2+10x-1 \\ 56=3x^2+10x-1 \\ 3x^2+10x-57=0 \end{gathered}$

Solve the quadratic equation

solve by using the formula

a=3

b=10

c=-57

substitute

$x=\frac{-10\pm\sqrt[\square]{10^2-4(3)(-57)}}{2(3)}$$x=\frac{-10\pm\sqrt[\square]{784}}{6}$$x=\frac{-10\pm28}{6}$

the values of x are

x=3 and x=-19/3