What is the critical f(x)=ln(x)+ln(64-x^2) ?

The critical f(x)=ln(x)+ln(64-x^2) is x= 8/(sqrt(3))

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critical f(x)=ln(x)+ln(64-x^2)

critical points

f (x) = ln (x) + ln (64 - x^{2})

x = \frac{8}{3}

Solution steps

ln (x) + ln (64 - x^{2})

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{8}{3}, x = \frac{8}{3}, x = - 8, x = 0, x = 8

Identify critical points not in

f (x)

domain

Domain of

ln (x) + ln (64 - x^{2}) : 0 < x < 8

ln (x) + ln (64 - x^{2})

is not defined at

x = - \frac{8}{3}, x = - 8, x = 0, x = 8

, therefore:

x = \frac{8}{3}

ln (1 - ln (x))

f (x) = x^{\frac{2}{3}} (x^{2} - 9)

f (x) = g (x) = 2 + (x - 5)^{3}

x^{3} + y^{3} + 3 x^{2} - 18 y + 5

1 - sin (x)

What is the critical f(x)=ln(x)+ln(64-x^2) ?
The critical f(x)=ln(x)+ln(64-x^2) is x= 8/(sqrt(3))