solvarmedw

2022-10-02

Increasing and decreasing intervals of a function
$f\left(x\right)={x}^{3}-4{x}^{2}+2$, which of the following statements are true:
(1) Increasing in $\left(-\mathrm{\infty },0\right)$, decreasing in $\left(\frac{8}{3},+\mathrm{\infty }\right)$.
(2) Increasing in both $\left(-\mathrm{\infty },0\right)$, decreasing in $\left(\frac{8}{3},+\mathrm{\infty }\right)$.
(3) decreasing in both $\left(-\mathrm{\infty },0\right)$, and $\left(\frac{8}{3},+\mathrm{\infty }\right)$.
(4) Decreasing in $\left(-\mathrm{\infty },0\right)$, Increasing in $\left(\frac{8}{3},+\mathrm{\infty }\right)$.
(5) None of the above.
${f}^{\prime }\left(x\right)=0=3{x}^{2}-8x=0⇒x=\frac{8}{3},x=0$ are the singular point/point of inflection.Could anyone tell me what next?

Paige Paul

Step 1
So, when you put ${f}^{\prime }\left(x\right)=0$ 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 $x<0$ and we will find that if ${x}_{1}<{x}_{2}$ then $f\left({x}_{1}\right) or we can see that ${f}^{\prime }\left(x\right)=3{x}^{2}-8x$ is positive for any $x<0$ and hence function is increasing in the interval $\left(-\mathrm{\infty },0\right]$.
Step 2
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 $x>8/3$

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