Question

let f(x)=x^3 - 6x^2 - 36x + 14. a. identify interval(s) where the graph of f(x) is increasing. (separate intervals with commas if needed.) b. identify interval(s) where the graph of f(x) is decreasing. (separate intervals with commas if needed.) c. find all relative maximum points. (enter ordered pair(s), separated by commas if needed. write none if there is no answer.) d. find all relative minimum points. (enter ordered pair(s), separated by commas if needed. write none if there is no answer.)

Explanation:

Step1: Find the derivative of $f(x)$

Using the power - rule $(x^n)'=nx^{n - 1}$, if $f(x)=x^{3}-6x^{2}-36x + 14$, then $f'(x)=3x^{2}-12x-36$.

Step2: Set $f'(x)=0$ to find critical points

$3x^{2}-12x - 36=0$. Divide through by 3: $x^{2}-4x - 12=0$. Factor the quadratic equation: $(x - 6)(x+2)=0$. So the critical points are $x=-2$ and $x = 6$.

Step3: Test intervals for increasing and decreasing

Choose test points in the intervals $(-\infty,-2)$, $(-2,6)$ and $(6,\infty)$.
For the interval $(-\infty,-2)$, let $x=-3$. Then $f'(-3)=3(-3)^{2}-12(-3)-36=27 + 36-36=27>0$, so $f(x)$ is increasing on $(-\infty,-2)$.
For the interval $(-2,6)$, let $x = 0$. Then $f'(0)=3(0)^{2}-12(0)-36=-36<0$, so $f(x)$ is decreasing on $(-2,6)$.
For the interval $(6,\infty)$, let $x = 7$. Then $f'(7)=3(7)^{2}-12(7)-36=147-84 - 36=27>0$, so $f(x)$ is increasing on $(6,\infty)$.

Step4: Find relative maximum and minimum

Since $f(x)$ changes from increasing to decreasing at $x=-2$, the relative maximum occurs at $x=-2$. $f(-2)=(-2)^{3}-6(-2)^{2}-36(-2)+14=-8-24 + 72+14=54$. The relative - maximum point is $(-2,54)$.
Since $f(x)$ changes from decreasing to increasing at $x = 6$, the relative minimum occurs at $x = 6$. $f(6)=(6)^{3}-6(6)^{2}-36(6)+14=216-216-216 + 14=-202$. The relative - minimum point is $(6,-202)$.

Answer:

a. $(-\infty,-2),(6,\infty)$
b. $(-2,6)$
c. $(-2,54)$
d. $(6,-202)$