for the following function, a) give the coordinates of any critical points and classify each point as a…

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

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

Answer

Explanation:

Step1: Find the first - derivative

Differentiate $k(x)=4x^{4}+16x^{3}$ using the power rule. The power rule states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. $k^\prime(x)=16x^{3}+48x^{2}=16x^{2}(x + 3)$

Step2: Find critical points

Set $k^\prime(x)=0$. $16x^{2}(x + 3)=0$ $x = 0$ or $x=-3$ When $x=-3$, $k(-3)=4(-3)^{4}+16(-3)^{3}=4\times81-16\times27=324 - 432=-108$ When $x = 0$, $k(0)=0$ So the critical points are $(-3,-108)$ and $(0,0)$

Step3: Classify critical points

Use the first - derivative test. Choose test points: Let's choose $x=-4$, $x=-1$ and $x = 1$ $k^\prime(-4)=16(-4)^{2}(-4 + 3)=16\times16\times(-1)<0$ $k^\prime(-1)=16(-1)^{2}(-1 + 3)=16\times1\times2>0$ $k^\prime(1)=16\times1^{2}(1 + 3)=16\times4>0$ Since $k^\prime(x)$ changes sign from negative to positive at $x=-3$, $(-3,-108)$ is a relative minimum. Since $k^\prime(x)$ does not change sign at $x = 0$, $(0,0)$ is neither a relative maximum nor a relative minimum.

Step4: Find the second - derivative

Differentiate $k^\prime(x)=16x^{3}+48x^{2}$ $k^{\prime\prime}(x)=48x^{2}+96x=48x(x + 2)$

Step5: Find points of inflection

Set $k^{\prime\prime}(x)=0$ $48x(x + 2)=0$ $x = 0$ or $x=-2$ When $x=-2$, $k(-2)=4(-2)^{4}+16(-2)^{3}=4\times16-16\times8=64 - 128=-64$ So the points of inflection are $(0,0)$ and $(-2,-64)$

Step6: Find intervals of concavity

Choose test points: Let's choose $x=-3$, $x=-1$ and $x = 1$ $k^{\prime\prime}(-3)=48(-3)(-3 + 2)=48\times(-3)\times(-1)>0$ $k^{\prime\prime}(-1)=48(-1)(-1 + 2)=48\times(-1)\times1<0$ $k^{\prime\prime}(1)=48\times1\times(1 + 2)=48\times3>0$ The function is concave up on $(-\infty,-2)\cup(0,\infty)$ and concave down on $(-2,0)$

Step7: Find intervals of increase and decrease

The function is decreasing on $(-\infty,-3)$ and increasing on $(-3,\infty)$

Answer:

a) Critical points: $(-3,-108)$ (relative minimum), $(0,0)$ (neither) b) Increasing on $(-3,\infty)$, decreasing on $(-\infty,-3)$ c) Points of inflection: $(0,0)$ and $(-2,-64)$ d) Concave up on $(-\infty,-2)\cup(0,\infty)$, concave down on $(-2,0)$ e) To sketch the graph: Mark the critical points $(-3,-108)$ and $(0,0)$, the points of inflection $(-2,-64)$ and $(0,0)$. The function is decreasing until $x=-3$, then increasing. It is concave up until $x=-2$, concave down from $x=-2$ to $x = 0$ and concave up again for $x>0$.