`f(x) = sqrt(x) - root(4)(x)` Find the local maximum and minimum values of `f` using both the First and Second Derivative Tests. Which method do you prefer?

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Chapter 4, 4.3 - Problem 21 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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marizi | High School Teacher | (Level 1) Associate Educator

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Using First Derivative Test, we follow:

 f'(a) >0  and f'(b) <0  in the interval a<c<b implies concave down and local maximum point occurs at x=c.

 f'(a) <0  and f'(b) >0  in the interval a<c<b implies concave up and local minimum point occurs at x=c.

Note:

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.

Sources:

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