# `G(x) = 5x^(2/3) - 2x^(5/3)` (a) FInd the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inflection points. (d)...

`G(x) = 5x^(2/3) - 2x^(5/3)` (a) FInd the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inflection points. (d) Use the information from parts (a)-(c) to sketch the graph. Check your work with a graphing device if you have one.

marizi | Certified Educator

a)  To find the intervals of increasing or decreasing f(x), recall:

--> f'(x) = positive value implies increasing f(x) of an interval I.

--> f'(x) = negative value implies decreasing f(x) of an interval I.

Applying power rule derivative on `f(x) = 5x^(2/3)-2x^(5/3)` :

`f'(x)= (2/3)*(5x^(2/3-1))-(5/3)*(2x^(5/3-1))`

`f'(x)= (10)/3x^(-1/3) -(5/3)*2x^(2/3)`

`f'(x)=(10)/(3x^(1/3))- (10x^(2/3))/3`

`f'(x)= (10)/(3x^(1/3)) -(10x^(2/3)*x^(1/3))/(3*x^(1/3))`

`f'(x)= (10 -10x)/(3x^(1/3)) `                     Note: `x^(2/3)*x^(1/3) = x^(2/3+1/3) or x^1`

Solve for critical value x=c by letting f'(x)=0 and D(x)=0.

Let D(x)=0:

`3x^(1/3)=0`

`3x^(1/3)*(1/3)=0*(1/3)`

`x^(1/3)=0`

`(x^(1/3))^3=0^3`

x=0

Let f'(x)=0:

`(10 -10x)/(3x^(1/3)) ` =0

`(10 -10x)= 0*(3x^(1/3))`

10-10x=0

10x=10

x=1

Table:

x          -1                0            0.5     1     2

f'(x)     -6.7     undefined         2.1    0     -2.6

intervals of decreasing f(x): (`-oo` ,0) and (1,`+ oo` )

interval of increasing f(x): (0,1)

b) Local extrema occurs at x=c when f'(c)=0.

f'(1)=0 then local extrema occurs at x=1.

Conditions:

`f'(a) gt0`  and `f'(b) lt0`  in the interval a<c<b implies concave down and local maximum point occurs at x=c.

`f'(a) lt0 `  and `f'(b) gt0`  in the interval a<c<b implies concave up and local minimum point occurs at x=c.

From the table, `f'(0.5) gt0 ` and` f'(2)lt0` then local maximum point occurs x=1.

Plug-in x=1 in `f(x)=5x^(2/3)-2x^(5/3).`

f(1)= 5(1)^(2/3)-2(1)^(5/3)

f(1)= 5-2

local maximum value: f(1)=3

c) According to second derivative test, we follow:

concave up when f"(c) >0 and concave down f"(c) >0 .

Inflection point occurs x=c when before and after x=c.

Applying power rule derivative on `f'(x)= (10 -10x)/(3x^(1/3)) ` or `f'(x)= (10x^(-1/3))/3 -(10x^(2/3))/3`

f"(x)`=(-1/3)*(10)/3x^(-1/3-1) -(2/3)*(10)/3x^(2/3-1)`

` `

f"(x)`=(-10)/9x^(-4/3) -(20)/9x^(-1/3)`

f"(x)`=-(10)/(9x^(4/3)) -(20)/(9x^(1/3))`

or `f"(x)=(-10-20x)/(9x^(4/3))`

Solve for inflection point:

`(-10-20x)/(9x^(4/3))=0`

`(-10-20x)=0* (9x^(4/3))`

-10-20x=0

20x=-10

x= `-1/2` or 0.5

Test of concavity:

x          -1         `-1/2 `        -1/3          1

f"(x)    `(10)/9`            0        -1.6

Intervals of concavity:

Concave up: (-`oo ` , -1/2)

Concave down: (-1/2, 0) and (0, +`oo` )

Inflection point occurs at `x=-1/2`

Plug-in x=-1/2 in f(x) =5x^(2/3)-2x^(5/3):

f(-1/2)=3.8

Inflection point (c, f(c)): (-0.5,3.8)

d) Graph: