Question

a. for the function and point below, find f(a). b. determine an equation of the line tangent to the graph of f at (a,f(a)) for the given value of a. f(x)= - 2x³, a = - 2 a. f(a)=□

Explanation:

Step1: Find the derivative of f(x)

Using the power - rule $(x^n)'=nx^{n - 1}$, if $f(x)=-2x^{3}$, then $f'(x)=-2\times3x^{2}=-6x^{2}$.

Step2: Evaluate f'(a)

Given $a = - 2$, substitute $x=-2$ into $f'(x)$. So $f'(-2)=-6\times(-2)^{2}=-6\times4=-24$.

Step3: Find f(a)

Substitute $x = - 2$ into $f(x)$. $f(-2)=-2\times(-2)^{3}=-2\times(-8)=16$.

Step4: Use the point - slope form for the tangent line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(a,f(a))=(-2,16)$ and $m = f'(a)=-24$.
So $y - 16=-24(x + 2)$.
Expand to get $y-16=-24x-48$, then $y=-24x - 32$.

Answer:

a. $f'(a)=-24$
b. $y=-24x - 32$