# Answer in Calculus for Josh #208334

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. ( )= 2 −4 / x²

2. =10 / 1+3 ²

3. = ³ − 3/2 ²

4. = − ³ / 3

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=-3sqrt[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 xin 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=pmdfrac{sqrt{3}}{3}”

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

“f(dfrac{sqrt{3}}{3})=dfrac{10(dfrac{sqrt{3}}{3})}{1+3(dfrac{sqrt{3}}{3})^2}=dfrac{5sqrt{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).”