What is the extreme f(x)=(25x^2+81)/x ?

The extreme f(x)=(25x^2+81)/x is Maximum(-9/5 ,-90),Minimum(9/5 ,90)

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extreme f(x)=(25x^2+81)/x

extreme

Maximum (- \frac{9}{5}, - 90), Minimum (\frac{9}{5}, 90)

Solution steps

f^{'} (x) = \frac{25 x ^{2} - 81}{x ^{2}}

Find intervals:

Increasing

: - \infty < x < - \frac{9}{5},

Decreasing

: - \frac{9}{5} < x < 0,

Decreasing

: 0 < x < \frac{9}{5},

Increasing

: \frac{9}{5} < x < \infty

Plug

x = - \frac{9}{5}

into

\frac{25 x ^{2} + 81}{x} : - 90

Maximum (- \frac{9}{5}, - 90)

Plug

x = \frac{9}{5}

into

\frac{25 x ^{2} + 81}{x} : 90

Minimum (\frac{9}{5}, 90)

Maximum (- \frac{9}{5}, - 90), Minimum (\frac{9}{5}, 90)

What is the extreme f(x)=(25x^2+81)/x ?
The extreme f(x)=(25x^2+81)/x is Maximum(-9/5 ,-90),Minimum(9/5 ,90)