Find the absolute maximum value and the absolute minimum value

Question

Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)

g(x)= 1/16x^2 - 16sqrt(x) on [0,25]

Max=

Min=

Answer 1

As the domain is [0, 25], we need to find where in the interior the first derivative is 0 along with the values at the limits.

g(x) = 1/16x^2 - 16 sqrt(x)

g'(x) = 1/8x -8/x^1/2

1/8 x = 8/x^1/2

Multiply both sides by 8 x ^1/2

x ^3/2 = 64

(x^3/2)^2/3 = 64^2/3 = (2⁶)^2/3 = 2⁴ = 16

x = 16

g''(x) = 1/8 + 4/x^3/2 is positive for x > 0 and continuous at 0.x = 16 is a relative minimum. Therefore, there is a minimum at 16. 1/16x^2 - 16 sqrt(x) = 1/16 * 16^2 - 16 * sqrt(16) = 16 - 16 * 4 = -48 is the minimum.

g(0) = 1/16 * 0^2 - 16 sqrt(0) = 0 - 0 = 0

g(25) = 1/16 * 25^2 - 16 sqrt(25) = 1/16*625 - 16 * 5 = 39 1/16 - 80 = - 40 15/16

g(0) > g(25)

The maximum is 0 at 0.