for the following exercise, use a graphing utility to estimate the local extrema of the function and to…

Explanation:

Step1: Find the derivative

Differentiate $f(x)=x^{4}-4x^{3}-3$ using the power - rule. The derivative $f^\prime(x)=4x^{3}-12x^{2}=4x^{2}(x - 3)$.

Step2: Find the critical points

Set $f^\prime(x)=0$. So $4x^{2}(x - 3)=0$. The solutions are $x = 0$ and $x=3$.

Step3: Test the intervals

Choose test points in the intervals $(-\infty,0)$, $(0,3)$ and $(3,\infty)$. For the interval $(-\infty,0)$, let $x=-1$. Then $f^\prime(-1)=4\times(-1)^{2}\times(-1 - 3)=4\times1\times(-4)=-16<0$, so the function is decreasing on $(-\infty,0)$. For the interval $(0,3)$, let $x = 1$. Then $f^\prime(1)=4\times1^{2}\times(1 - 3)=4\times1\times(-2)=-8<0$, so the function is decreasing on $(0,3)$. For the interval $(3,\infty)$, let $x = 4$. Then $f^\prime(4)=4\times4^{2}\times(4 - 3)=4\times16\times1 = 64>0$, so the function is increasing on $(3,\infty)$. Since the function changes from decreasing to increasing at $x = 3$, there is a local minimum at $x = 3$. $f(3)=3^{4}-4\times3^{3}-3=81-108 - 3=-30$.

Answer:

Local extrema: $(3,-30)$ Increasing on the interval: $(3,\infty)$ Decreasing on the interval: $(-\infty,3)$