Question

hw 15 - increasing and decreasing functions section 3.1: problem 5 (1 point)
the function
f(x)=2x^{3}-3x^{2}-180x + 4
is decreasing on the interval ( , ).
it is increasing on the interval (-\infty, )
and the interval ( , \infty ).

Explanation:

Step1: Find the derivative

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $f'(x)=6x^{2}-6x - 180$.

Step2: Set the derivative equal to zero

$6x^{2}-6x - 180 = 0$. Divide through by 6: $x^{2}-x - 30=0$.

Step3: Factor the quadratic equation

$(x - 6)(x + 5)=0$. So the critical points are $x=-5$ and $x = 6$.

Step4: Test intervals

Choose test points in the intervals $(-\infty,-5)$, $(-5,6)$ and $(6,\infty)$. Let's choose $x=-6$, $x = 0$ and $x=7$.
For $x=-6$, $f'(-6)=6\times(-6)^{2}-6\times(-6)-180=216 + 36-180=72>0$.
For $x = 0$, $f'(0)=6\times0^{2}-6\times0 - 180=-180<0$.
For $x=7$, $f'(7)=6\times7^{2}-6\times7 - 180=294-42 - 180=72>0$.

Answer:

The function is decreasing on the interval $(-5,6)$. It is increasing on the interval $(-\infty,-5)$ and the interval $(6,\infty)$.