What is the extreme f(x)=(sqrt(5-x^2))/x ?

The extreme f(x)=(sqrt(5-x^2))/x is Maximum(-sqrt(5),0),Minimum(sqrt(5),0)

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extreme f(x)=(sqrt(5-x^2))/x

extreme points

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

Maximum (- 5, 0), Minimum (5, 0)

Solution steps

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

Find intervals:

Decreasing

: - 5 < x < 0,

Decreasing

: 0 < x < 5

Plug

x = - 5

into

\frac{5 - x ^{2}}{x} : 0

Maximum (- 5, 0)

Plug

x = 5

into

\frac{5 - x ^{2}}{x} : 0

Minimum (5, 0)

Maximum (- 5, 0), Minimum (5, 0)

x^{2} - 14 x + 2

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

y = 2 x^{2} - 5 x + 4 - 10

x^{4} - 4 x^{3} - 18 x^{2} - 20 x - 12

- 0.015 x^{2} + 1.24 x - 7.4

What is the extreme f(x)=(sqrt(5-x^2))/x ?
The extreme f(x)=(sqrt(5-x^2))/x is Maximum(-sqrt(5),0),Minimum(sqrt(5),0)