Question

use a graphing utility to graph the function and visually determine the intervals over which the function is increasing, decreasing, or constant. $f(x)=-x^{3}+3x + 1$

Explanation:

Step1: Find the derivative

Differentiate $f(x)=-x^{3}+3x + 1$ using the power - rule. The derivative $f'(x)=-3x^{2}+3$.

Step2: Set the derivative equal to zero

Solve $-3x^{2}+3 = 0$. First, factor out - 3: $-3(x^{2}-1)=0$, then $x^{2}-1=(x - 1)(x + 1)=0$. So $x=-1$ or $x = 1$.

Step3: Test intervals

Choose test points in the intervals $(-\infty,-1)$, $(-1,1)$ and $(1,\infty)$.
For the interval $(-\infty,-1)$, let $x=-2$. Then $f'(-2)=-3\times(-2)^{2}+3=-12 + 3=-9<0$, so the function is decreasing on $(-\infty,-1)$.
For the interval $(-1,1)$, let $x = 0$. Then $f'(0)=-3\times0^{2}+3=3>0$, so the function is increasing on $(-1,1)$.
For the interval $(1,\infty)$, let $x = 2$. Then $f'(2)=-3\times2^{2}+3=-12 + 3=-9<0$, so the function is decreasing on $(1,\infty)$.

Answer:

The function $f(x)=-x^{3}+3x + 1$ is increasing on the interval $(-1,1)$, and decreasing on the intervals $(-\infty,-1)$ and $(1,\infty)$.