What is the extreme f(x)=((x^2+121))/(2x) ?

The extreme f(x)=((x^2+121))/(2x) is Maximum(-11,-11),Minimum(11,11)

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extreme f(x)=((x^2+121))/(2x)

extreme points

f (x) = \frac{( x ^{2} + 121 )}{2 x}

Maximum (- 11, - 11), Minimum (11, 11)

Solution steps

f^{'} (x) = \frac{x ^{2} - 121}{2 x ^{2}}

Find intervals:

Increasing

: - \infty < x < - 11,

Decreasing

: - 11 < x < 0,

Decreasing

: 0 < x < 11,

Increasing

: 11 < x < \infty

Plug

x = - 11

into

\frac{x ^{2} + 121}{2 x} : - 11

Maximum (- 11, - 11)

Plug

x = 11

into

\frac{x ^{2} + 121}{2 x} : 11

Minimum (11, 11)

Maximum (- 11, - 11), Minimum (11, 11)

f (x) = - 7 + 12 x - 3 x^{2}

2 x^{3} - 6

f (x) = (x^{2} - 7) (x^{2} - 6) (x^{2} + 4)

f (x) = (x - 3)^{\frac{4}{3}}, - 7 \leq x \leq 7

2 x^{3} - 2 x^{2} - x - 9

What is the extreme f(x)=((x^2+121))/(2x) ?
The extreme f(x)=((x^2+121))/(2x) is Maximum(-11,-11),Minimum(11,11)