Question

Given the function f, discuss its relative maximum and minimum

points, the intervals where it is increasing and decreasing, the intervals of concavity, and the points of inflection. Construct a sketch of the graph of the function.

1. 𝑓(𝑥)= 2𝑥−4 / x²

2. 𝑓 𝑥 =10𝑥 / 1+3𝑥²

3. 𝑓 𝑥 = 𝑥³ − 3/2 𝑥²

4. 𝑓 𝑥 = 𝑥 − 𝑥³ / 3

Answer 1

1. $f(x)=2x-4/x^2$

$Df: (-\infin, 0)\cup(0, \infin)$

$$f'(x)=(2x-\dfrac{4}{x^2})'=2+\dfrac{8}{x^3}$$

Critical number(s):

$$f'(x)=0=>2+\dfrac{8}{x^3}=0=>x=-\sqrt[3]{4}$$

$$f(-\sqrt[3]{4})=2(-\sqrt[3]{4})-4/(-\sqrt[3]{4})^2=-3\sqrt[3]{4}$$

If $x<-\sqrt[3]{4}, f'(x)>0, f(x)$ increases.

If $-\sqrt[3]{4}<x<0, f'(x)<0, f(x)$ decreases.

If $x>0, f'(x)>0, f(x)$ increases.

The function $f(x)$ increases on $(-\infin, -\sqrt[3]{4})\cup(0, \infin).$

The function $f(x)$ decreases on $(-\sqrt[3]{4}, 0).$

The function $f(x)$ has a local maximum at $x=-\sqrt[3]{4}.$

The function $f(x)$ has no local minimum.

$$f''(x)=(2+\dfrac{8}{x^3})'=-\dfrac{32}{x^4}<0 x\in Df$$

The function $f(x)$ is concave down on $(-\infin, 0)\cup(0, \infin).$

The function $f(x)$ is not concave up anywhere.

The function $f(x)$ has no point of inflection.

2. $f(x)=\dfrac{10x}{1+3x^2}$

$Df: (-\infin, \infin)$

$$f'(x)=(\dfrac{10x}{1+3x^2})'=-\dfrac{10(3x^2-1)}{(1+3x^2)^2}$$

Critical number(s):

$$f'(x)=0=>-\dfrac{10(3x^2-1)}{(1+3x^2)^2}=0=>x=\pm\dfrac{\sqrt{3}}{3}$$

$$f(-\dfrac{\sqrt{3}}{3})=\dfrac{10(-\dfrac{\sqrt{3}}{3})}{1+3(-\dfrac{\sqrt{3}}{3})^2}=-\dfrac{5\sqrt{3}}{3}$$

$$f(\dfrac{\sqrt{3}}{3})=\dfrac{10(\dfrac{\sqrt{3}}{3})}{1+3(\dfrac{\sqrt{3}}{3})^2}=\dfrac{5\sqrt{3}}{3}$$

If $x<-\dfrac{\sqrt{3}}{3}, f'(x)<0, f(x)$ decreases.

If $-\dfrac{\sqrt{3}}{3}<x<\dfrac{\sqrt{3}}{3}, f'(x)>0, f(x)$ increases.

If $x>\dfrac{\sqrt{3}}{3}, f'(x)<0, f(x)$ decreases.

The function $f(x)$ increases on $(-\dfrac{\sqrt{3}}{3}, \dfrac{\sqrt{3}}{3}).$

The function $f(x)$ decreases on $(-\infin, -\dfrac{\sqrt{3}}{3})\cup(\dfrac{\sqrt{3}}{3}, \infin).$

The function $f(x)$ has a local maximum at $x=\dfrac{\sqrt{3}}{3}.$

The function $f(x)$ has a local minimum at $x=-\dfrac{\sqrt{3}}{3}.$

$$f''(x)=(-\dfrac{10(3x^2-1)}{(1+3x^2)^2})'=\dfrac{180x(x^2-1)}{(1+3x^2)^3}$$

$$f''(x)=0=>\dfrac{180x(x^2-1)}{(1+3x^2)^3}=0$$

$$x_1=-1, x_2=0, x_3=1$$

$$f(-1)=\dfrac{10(-1)}{1+3(-1)^2}=-\dfrac{5}{2}$$

$$f(0)=\dfrac{10(0)}{1+3(0)^2}=0$$

$$f(1)=\dfrac{10(1)}{1+3(1)^2}=\dfrac{5}{2}$$

The function $f(x)$ is concave up on $(-1, 0)\cup(1, \infin).$

The function $f(x)$ is concave down on $(-\infin, -1)\cup(0, 1).$

The points of inflection are $(-1, -2.5), (0,0), (1,2.5).$

3. $f(x)=x^3-\dfrac{3}{2x^2}$

$Df: (-\infin, 0)\cup(0, \infin)$

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

Critical number(s):

$$f'(x)=0=>3x^2+\dfrac{3}{x^3}=0=>x=-1$$

$$f(-1)=(-1)^3-\dfrac{3}{2(-1)^2}=-\dfrac{5}{2}$$

If $x<-1, f'(x)>0, f(x)$ increases.

If $-1<x<0, f'(x)<0, f(x)$ decreases.

If $x>0, f'(x)>0, f(x)$ increases.

The function $f(x)$ increases on $(-\infin, -1)\cup(0, \infin).$

The function $f(x)$ decreases on $(-1, 0).$

The function $f(x)$ has a local maximum at $x=-1.$

The function $f(x)$ has no local minimum.

$$f''(x)=(3x^2+\dfrac{3}{x^3})'=6x-\dfrac{9}{x^4}$$ $$f''(x)=0=>6x-\dfrac{9}{x^4}=0$$

$$x=\sqrt[5]{1.5}$$

$$f(\sqrt[5]{1.5})=(\sqrt[5]{1.5})^3-\dfrac{3}{2(\sqrt[5]{1.5})^2}=0$$

The function $f(x)$ is concave up on $(\sqrt[5]{1.5}, \infin).$

The function $f(x)$ is concave down on $(-\infin, 0)\cup(0, \sqrt[5]{1.5}).$

The point of inflection is $(\sqrt[5]{1.5},0).$

4. $f(x)=x-\dfrac{x^3}{3}$

$Df: (-\infin, \infin)$

$$f'(x)=(x-\dfrac{x^3}{3})'=1-x^2$$

Critical number(s):

$$f'(x)=0=>1-x^2=0=>x=\pm1$$

$$f(-1)=-1-\dfrac{(-1)^3}{3}=-\dfrac{2}{3}$$

$$f(1)=1-\dfrac{(1)^3}{3}=\dfrac{2}{3}$$

If $x<-1, f'(x)<0, f(x)$ decreases.

If $-1<x<1, f'(x)>0, f(x)$ increases.

If $x>1, f'(x)<0, f(x)$ decreases.

The function $f(x)$ increases on $(-\dfrac{\sqrt{3}}{3}, \dfrac{\sqrt{3}}{3}).$

The function $f(x)$ decreases on $(-\infin, -\dfrac{\sqrt{3}}{3})\cup(\dfrac{\sqrt{3}}{3}, \infin).$

The function $f(x)$ has a local maximum at $x=1.$

The function $f(x)$ has a local minimum at $x=-1.$

$$f''(x)=(1-x^2)'=-2x$$

$$f''(x)=0=>-2x=0$$

$$x=0$$

$$f(0)=0-\dfrac{(0)^3}{3}=0$$

The function $f(x)$ is concave up on $(-\infin, 0).$

The function $f(x)$ is concave down on $(0,-\infin).$

The point of inflection is $(0,0).$