Question

throughout which of the following intervals is $f(x)=-x^{3}+2x^{2}+4x - 2$ only decreasing?
$-infty$-infty$1$2

Explanation:

Step1: Find the derivative of f(x)

$f'(x)=-3x^{2}+4x + 4$

Step2: Set the derivative equal to zero to find critical points

$-3x^{2}+4x + 4 = 0$
$3x^{2}-4x - 4=0$
$(3x + 2)(x - 2)=0$
The critical points are $x=-\frac{2}{3}$ and $x = 2$.

Step3: Determine the sign of the derivative in intervals

We consider the intervals $(-\infty,-\frac{2}{3})$, $(-\frac{2}{3},2)$ and $(2,\infty)$.
For $x\in(-\infty,-\frac{2}{3})$, let's take $x=-1$, then $f'(-1)=-3-4 + 4=-3<0$.
For $x\in(-\frac{2}{3},2)$, let's take $x = 0$, then $f'(0)=4>0$.
For $x\in(2,\infty)$, let's take $x = 3$, then $f'(3)=-3\times9+4\times3 + 4=-27 + 12+4=-11<0$.
The function $f(x)$ is decreasing when $f'(x)<0$, which is for $x\in(-\infty,-\frac{2}{3})\cup(2,\infty)$. The interval $2 < x<4$ is a sub - interval where the function is only decreasing.

Answer:

$2 < x<4$