Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 40

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 af'(a) lt0 and f'(b) gt0 in the interval aFrom 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:

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