What is the extreme f(x)=(3-sqrt(9-x^2))/x ?

The extreme f(x)=(3-sqrt(9-x^2))/x is Minimum(-3,-1),Maximum(3,1)

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extreme f(x)=(3-sqrt(9-x^2))/x

extreme

Minimum (- 3, - 1), Maximum (3, 1)

Solution steps

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

Find intervals:

Increasing

: - 3 < x < 0,

Increasing

: 0 < x < 3

Plug

x = - 3

into

\frac{3 - 9 - x ^{2}}{x} : - 1

Minimum (- 3, - 1)

Plug

x = 3

into

\frac{3 - 9 - x ^{2}}{x} : 1

Maximum (3, 1)

Minimum (- 3, - 1), Maximum (3, 1)

What is the extreme f(x)=(3-sqrt(9-x^2))/x ?
The extreme f(x)=(3-sqrt(9-x^2))/x is Minimum(-3,-1),Maximum(3,1)