$f(x)= x^3-4x^2+2$, which of the following statements are true:
(1) Increasing in $(-\infty, 0)$, decreasing in $(\frac{8}{3}, +\infty)$
(2) Increasing in both $(-\infty, 0)$, and $(\frac{8}{3}, +\infty)$
(3) decreasing in both $(-\infty, 0)$, and $(\frac{8}{3}, +\infty)$
(4) Decreasing in $(-\infty, 0)$, Increasing in $(\frac{8}{3}, +\infty)$
(5) None of the above.
$f'(x)=0=3x^2-8x=0 \Rightarrow x=\frac{8}{3}, x=0$ are the singular point/point of inflection.Could anyone tell me what next?
So, when you put $f’(x)=0$ and got $x=0, 8/3$, it means that function takes the “u-turn” at those points. Now, we need to check what was happening before $x=0$, what’s happening between $0$ and $8/3$ and what will happen after $x=8/3$.
We can take any two $x$’s such that $x \lt 0$ and we will find that if $x_1 \lt x_2$ then $f(x_1) \lt f(x_2)$ or we can see that $f’(x)= 3x^2 -8x$ is positive for any $x \lt 0$ and hence function is increasing in the interval $(-\infty, 0]$.
And as we know that the function will take a “u-turn” at $x=0$ so, the function will decrease between $0$and $8/3$ and again after a “u-turn” at $x=8/3$ it will increase. If you find this “u-turn” concept informal you can go for similar method above and, you will find that $f$ is decreasing between $x=0$ and $x=8/3$, and finally it is increasing for $x \gt 8/3$.
Hope it helps!