Almintas2l

2022-07-23

Finding f intervals

How would I find the f intervals for the following two functions.

$f(x)=(x-2{)}^{2}(x+1{)}^{2}$

using the chain rule I got $(x-2{)}^{2}(2)(x+1)(1)+(2)(x-2)(1)(x+1{)}^{2}$

then I got f decrease $(-\mathrm{\infty},-1]$

and f increase $[2,\mathrm{\infty})$

but the area between -1 and 2 in confusing me.

My second function is $f(x)=x+\frac{4}{{x}^{2}}$

differentiating I got $\frac{{x}^{3}-8}{{x}^{3}}$

so I got f increase $(\mathrm{\infty},0)$ and $[2,\mathrm{\infty})$

f decrease (0, 2]

but did I do this correctly.

How would I find the f intervals for the following two functions.

$f(x)=(x-2{)}^{2}(x+1{)}^{2}$

using the chain rule I got $(x-2{)}^{2}(2)(x+1)(1)+(2)(x-2)(1)(x+1{)}^{2}$

then I got f decrease $(-\mathrm{\infty},-1]$

and f increase $[2,\mathrm{\infty})$

but the area between -1 and 2 in confusing me.

My second function is $f(x)=x+\frac{4}{{x}^{2}}$

differentiating I got $\frac{{x}^{3}-8}{{x}^{3}}$

so I got f increase $(\mathrm{\infty},0)$ and $[2,\mathrm{\infty})$

f decrease (0, 2]

but did I do this correctly.

Bianca Chung

Beginner2022-07-24Added 16 answers

Step 1

As you did f′(x) correctly, we have ${f}^{\prime}(x)=2(x-2)(x+1)(2x-1)$

Step 2

When:

- $x\le -1\text{}\text{}$, $(x-2)\le 0$ and then two terms $(x+1)$ and $(2x-1)$ are negative. So f′(x) is negative.

- $x\le \frac{1}{2}\text{}\text{}$, $(x-2)\le 0$ and $(x+1)>0$ and $(2x-1)<0$ is negative, so f′(x) is positive.

- $x\le 2\text{}\text{}$, $(x-2)\le 0$ and then two terms $(x+1)$ and $(2x-1)$ are positive, so f′(x) is nagative again.

- And if $x>2$ all terms are positive and so f′(x) is positive again.

As you did f′(x) correctly, we have ${f}^{\prime}(x)=2(x-2)(x+1)(2x-1)$

Step 2

When:

- $x\le -1\text{}\text{}$, $(x-2)\le 0$ and then two terms $(x+1)$ and $(2x-1)$ are negative. So f′(x) is negative.

- $x\le \frac{1}{2}\text{}\text{}$, $(x-2)\le 0$ and $(x+1)>0$ and $(2x-1)<0$ is negative, so f′(x) is positive.

- $x\le 2\text{}\text{}$, $(x-2)\le 0$ and then two terms $(x+1)$ and $(2x-1)$ are positive, so f′(x) is nagative again.

- And if $x>2$ all terms are positive and so f′(x) is positive again.

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