What is the extreme f(x)=(sqrt(3-x^2))/x ?

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

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extreme f(x)=(sqrt(3-x^2))/x

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3

Plug

x = - 3

into

\frac{3 - x ^{2}}{x} : 0

Maximum (- 3, 0)

Plug

x = 3

into

\frac{3 - x ^{2}}{x} : 0

Minimum (3, 0)

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

f (x) = 3 e^{\frac{x}{2}} - (x + 1) x + 1

f (x) = \frac{e ^{x}}{9 + e ^{x}}

6 x y - 1.2 x^{2} - 3 y^{4}

f (x) = x^{\frac{1}{2}} (x - 4)

\frac{2 x + 1}{x - 3}

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