What is the extreme f(x)=(sqrt(9-x^2))/x ?

The extreme f(x)=(sqrt(9-x^2))/x is Maximum(-3,0),Minimum(3,0)

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extreme f(x)=(sqrt(9-x^2))/x

extreme points

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

Maximum (- 3, 0), Minimum (3, 0)

Solution steps

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

Find intervals:

Decreasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3

Plug

x = - 3

into

\frac{9 - x ^{2}}{x} : 0

Maximum (- 3, 0)

Plug

x = 3

into

\frac{9 - x ^{2}}{x} : 0

Minimum (3, 0)

Maximum (- 3, 0), Minimum (3, 0)

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

f (x) = e^{x - y} (x^{2} - 2 y^{2})

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

f (x) = \frac{3}{4} x^{4} - 3 x^{3} - 81 x^{2} - 16

f (x) = 3 + 81 x - 3 x^{3}

What is the extreme f(x)=(sqrt(9-x^2))/x ?
The extreme f(x)=(sqrt(9-x^2))/x is Maximum(-3,0),Minimum(3,0)