The value of x which S(x) is a global minimum is x = 1/2
From the question, we have:
This means that:
S(x) = 1/x + 4x^2
From the question, we understand that x is a positive number.
This means that the domain of x is x > 0
As a notation, we have (0, ∞)
Recall that:
S(x) = 1/x + 4x^2
Differentiate the function
S'(x) = -1/x^2 + 8x
Set to 0
-1/x^2 + 8x = 0
Multiply through by x^2
-1 + 8x^3 = 0
Add 1 to both sides
8x^3 = 1
Divide by 8
x^3 = 1/8
Take the cube root of both sides
x = 1/2
To prove the point is a global minimum, we have:
S'(x) = -1/x^2 + 8x
Determine the second derivative
S''(x) = 2/x^3 + 8
Set x = 1/2
S''(x) = 2/(1/2)^3 + 8
Evaluate the exponent
S''(x) = 2/1/8 + 8
Evaluate the quotient
S''(x) = 16 + 8
Evaluate the sum
S''(x) = 24
Because S'' is positive, then the single critical point is a global minimum
Read more about second derivative test at:
https://brainly.com/question/14261130
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