Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 21

Using First Derivative Test, we follow:
f'(a) >0 and f'(b) <0 in the interval a f'(a) <0 and f'(b) >0 in the interval aNote:
When f'(x) >0 or f'(x) = positive value then function f(x) is increasing or has a positive slope of tangent line (slant line going up).
When f'(x)< 0 or f'(x) = negative value then function f(x) is decreasing or has a negative slope of tangent line (slant line going down).
Applying power rule derivative on f(x) = x^(1/2) -x^(1/4):
f'(x)=(1/2)x^(1/2-1) -(1/4)x^(1/4-1)
f'(x)=(1/2)x^(-1/2) -(1/4)x^(-3/4)
f'(x)=(1/(2x^(1/2))) *((2x^(1/4))/(2x^(1/4)))-1/(4x^(3/4))
f'(x)=(2x^(1/4) -1)/(4x^(3/4))
Let f'(x)=0:
(2x^(1/4) -1)/(4x^(3/4))=0
2x^(1/4) -1=0*(4x^(3/4))
2x^(1/4) -1=0
2x^(1/4) = 1
x^(1/4) = 1/2
(x^(1/4))^4 = (1/2)^4
x=1/(16)or 0.0625
Table:
x 0.04 0.0625 0.08
f'(x) -0.295 0 0.106
f(x) decreasing increasing
It shows f'(0.04) <0 and f'(0.08)> 0 then local minimum point occurs at x=1/(16) or 0.0625.
As for Second Derivative Test, a critical point at x=c such that f'(c) =0 and f"(c) is continuous around the region of x=c follows:
f"(c) >0 then local maximum occurs at x=c.
f"(c) <0 then local minimum occurs at x=c
Applying power rule on f'(x)=(1/2)x^(-1/2) -(1/4)x^(-3/4) :
f"(x) =(-1/2)(1/2)x^(-1/2-1) -(-3/4)(1/4)x^(-3/4-1)
=(-1/4) x^(-3/2) +3/(16) x^(-7/4)
=-1/(4x^(3/2)) +3/(16x^(7/4))
Plug-in x=1/(16) in f"(x)=-1/(4x^(3/2)) +3/(16x^(7/4)).
f"(1/(16) )= 8 positive value
Continuation....
For the actual local minimum value of f, we plug-in x= 1/(16) in f(x)=x^(1/2)-x^(1/4) :
f(1/(16)) =(1/(16))^(1/2)-(1/(16))^(1/4)
f(1/(16)) = 1/4- 1/2
f(1/(16))= -1/4 or -0.25
or f"(1/(16) )> 0 then local minimum occurs at x=1/(16)
First derivative test will do if the second derivative function is not easy to derived.
Second derivative test will be easier if f"(x) is solvable with few steps.
This way we can plug-in critical value x=c directly on f"(x) to check of the local extrema is maximum or minimum.