Question

for the following function, a) give the coordinates of any critical points and classify each point as a relative maximum, a relative minimum, or neither, b) identify intervals where the function is increasing or decreasing, c) give the coordinates of any points of inflection, d) identify intervals where the function is concave up or concave down, and e) sketch the graph. k(x)=4x^4 + 16x^3. a) what are the coordinates of the relative extrema? select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. a. the relative minimum point(s) is/are and the relative maximum point(s) is/are (simplify your answers. use integers or fractions for any numbers in the expression. type an ordered - pair. use a comma to separate answers as needed.) b. the relative maximum point(s) is/are and there are no relative minimum point(s) (simplify your answer. use integers or fractions for any numbers in the expression. type an ordered - pair. use a comma to separate answers as needed.) c. the relative minimum point(s) is/are and there are no relative maximum point(s) (simplify your answer. use integers or fractions for any numbers in the expression. type an ordered - pair. use a comma to separate answers as needed.) d. there are no relative minimum points and there are no relative maximum points.

Explanation:

Step1: Find the first - derivative

Differentiate $k(x)=4x^{4}+16x^{3}$ using the power rule $\frac{d}{dx}(x^{n}) = nx^{n - 1}$.
$k'(x)=16x^{3}+48x^{2}=16x^{2}(x + 3)$

Step2: Find the critical points

Set $k'(x)=0$.
$16x^{2}(x + 3)=0$
$x = 0$ or $x=-3$

Step3: Use the first - derivative test

Choose test points in the intervals $(-\infty,-3)$, $(-3,0)$ and $(0,\infty)$.
For $x=-4$ (in $(-\infty,-3)$), $k'(-4)=16\times(-4)^{2}\times(-4 + 3)=16\times16\times(-1)<0$.
For $x=-1$ (in $(-3,0)$), $k'(-1)=16\times(-1)^{2}\times(-1 + 3)=16\times1\times2>0$.
For $x = 1$ (in $(0,\infty)$), $k'(1)=16\times1^{2}\times(1 + 3)=16\times1\times4>0$.
Since the function changes from decreasing to increasing at $x=-3$, $(-3,k(-3))$ is a relative minimum.
$k(-3)=4\times(-3)^{4}+16\times(-3)^{3}=4\times81-16\times27=324 - 432=-108$.
The relative minimum point is $(-3,-108)$ and there are no relative maximum points.

Answer:

C. The relative minimum point(s) is/are $(-3,-108)$ and there are no relative maximum point(s)