hw 15 - increasing and decreasing functions section 3.1: pr(1 point)\nconsider the function\nf(x)=x^{3}+6x^{2…

hw 15 - increasing and decreasing functions section 3.1: pr(1 point)\nconsider the function\nf(x)=x^{3}+6x^{2}-13.\na) find the critical numbers of (f). (separate multiple answers by commas.)\nb) determine the intervals on which (f) is increasing and decreasing.\nf is increasing on:\nf is decreasing on:

hw 15 - increasing and decreasing functions section 3.1: pr(1 point)\nconsider the function\nf(x)=x^{3}+6x^{2}-13.\na) find the critical numbers of (f). (separate multiple answers by commas.)\nb) determine the intervals on which (f) is increasing and decreasing.\nf is increasing on:\nf is decreasing on:

Answer

Explanation:

Step1: Find the derivative

Differentiate $f(x)=x^{3}+6x^{2}-13$ using the power - rule. The derivative $f'(x)=3x^{2}+12x$.

Step2: Find critical numbers

Set $f'(x) = 0$. So, $3x^{2}+12x=3x(x + 4)=0$. Solving $3x(x + 4)=0$ gives $x=0$ and $x=-4$.

Step3: Determine intervals of increase and decrease

We use test points in the intervals $(-\infty,-4)$, $(-4,0)$ and $(0,\infty)$. For the interval $(-\infty,-4)$, let $x=-5$. Then $f'(-5)=3\times(-5)^{2}+12\times(-5)=75 - 60 = 15>0$, so $f(x)$ is increasing on $(-\infty,-4)$. For the interval $(-4,0)$, let $x=-1$. Then $f'(-1)=3\times(-1)^{2}+12\times(-1)=3 - 12=-9<0$, so $f(x)$ is decreasing on $(-4,0)$. For the interval $(0,\infty)$, let $x = 1$. Then $f'(1)=3\times1^{2}+12\times1=3 + 12 = 15>0$, so $f(x)$ is increasing on $(0,\infty)$.

Answer:

a) $-4,0$ b) $f$ is increasing on: $(-\infty,-4)\cup(0,\infty)$ $f$ is decreasing on: $(-4,0)$