Answer to Question #208334 in Calculus for Josh

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 $(0,0).$