solvarmedw

2022-10-02

Increasing and decreasing intervals of a function

$f(x)={x}^{3}-4{x}^{2}+2$, which of the following statements are true:

(1) Increasing in $(-\mathrm{\infty},0)$, decreasing in $(\frac{8}{3},+\mathrm{\infty})$.

(2) Increasing in both $(-\mathrm{\infty},0)$, decreasing in $(\frac{8}{3},+\mathrm{\infty})$.

(3) decreasing in both $(-\mathrm{\infty},0)$, and $(\frac{8}{3},+\mathrm{\infty})$.

(4) Decreasing in $(-\mathrm{\infty},0)$, Increasing in $(\frac{8}{3},+\mathrm{\infty})$.

(5) None of the above.

${f}^{\prime}(x)=0=3{x}^{2}-8x=0\Rightarrow x=\frac{8}{3},x=0$ are the singular point/point of inflection.Could anyone tell me what next?

$f(x)={x}^{3}-4{x}^{2}+2$, which of the following statements are true:

(1) Increasing in $(-\mathrm{\infty},0)$, decreasing in $(\frac{8}{3},+\mathrm{\infty})$.

(2) Increasing in both $(-\mathrm{\infty},0)$, decreasing in $(\frac{8}{3},+\mathrm{\infty})$.

(3) decreasing in both $(-\mathrm{\infty},0)$, and $(\frac{8}{3},+\mathrm{\infty})$.

(4) Decreasing in $(-\mathrm{\infty},0)$, Increasing in $(\frac{8}{3},+\mathrm{\infty})$.

(5) None of the above.

${f}^{\prime}(x)=0=3{x}^{2}-8x=0\Rightarrow x=\frac{8}{3},x=0$ are the singular point/point of inflection.Could anyone tell me what next?

Paige Paul

Beginner2022-10-03Added 11 answers

Step 1

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

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 $x>8/3$

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

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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