What is the extreme f(x)=((x-6))/(x^2) ?

The extreme f(x)=((x-6))/(x^2) is Maximum(12, 1/24)

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extreme f(x)=((x-6))/(x^2)

extreme points

f (x) = \frac{( x - 6 )}{x ^{2}}

Maximum (12, \frac{1}{24})

Solution steps

f^{'} (x) = \frac{- x + 12}{x ^{3}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 12,

Decreasing

: 12 < x < \infty

Plug

x = 12

into

\frac{x - 6}{x ^{2}} : \frac{1}{24}

Maximum (12, \frac{1}{24})

f (x) = (x - 2)^{3} + 8

f (x) = x^{\frac{5}{3}} - 2 x^{\frac{2}{3}}

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

f (x, y) = x^{2} + y^{2} - 4 x y

f (x) = ∣ x ∣ + \frac{x}{7}

What is the extreme f(x)=((x-6))/(x^2) ?
The extreme f(x)=((x-6))/(x^2) is Maximum(12, 1/24)