Question

answer the questions for the function f(x)= -3x^3 + 3x^2 - x - 5
a. find formulas for f(x) and f(x).
f(x)=
f(x)=
enter f(x), f(x), and f(x) into your grapher to examine the table.

Explanation:

Step1: Apply power - rule for first - derivative

The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$. For $f(x)=-3x^{3}+3x^{2}-x - 5$, we have:
$f^\prime(x)=\frac{d}{dx}(-3x^{3})+\frac{d}{dx}(3x^{2})+\frac{d}{dx}(-x)+\frac{d}{dx}(-5)$.
Using the power - rule: $\frac{d}{dx}(-3x^{3})=-3\times3x^{3 - 1}=-9x^{2}$, $\frac{d}{dx}(3x^{2})=3\times2x^{2 - 1}=6x$, $\frac{d}{dx}(-x)=-1$, and $\frac{d}{dx}(-5)=0$.
So, $f^\prime(x)=-9x^{2}+6x - 1$.

Step2: Apply power - rule for second - derivative

Differentiate $f^\prime(x)=-9x^{2}+6x - 1$ with respect to $x$.
$f^{\prime\prime}(x)=\frac{d}{dx}(-9x^{2})+\frac{d}{dx}(6x)+\frac{d}{dx}(-1)$.
Using the power - rule: $\frac{d}{dx}(-9x^{2})=-9\times2x^{2 - 1}=-18x$, $\frac{d}{dx}(6x)=6$, and $\frac{d}{dx}(-1)=0$.
So, $f^{\prime\prime}(x)=-18x + 6$.

Answer:

$f^\prime(x)=-9x^{2}+6x - 1$
$f^{\prime\prime}(x)=-18x + 6$