What is the extreme f(x)=(1+x^2)/x ?

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

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extreme f(x)=(1+x^2)/x

extreme points

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

Maximum (- 1, - 2), Minimum (1, 2)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

\frac{1 + x ^{2}}{x} : - 2

Maximum (- 1, - 2)

Plug

x = 1

into

\frac{1 + x ^{2}}{x} : 2

Minimum (1, 2)

Maximum (- 1, - 2), Minimum (1, 2)

f (x, y) = 2 x^{2} + 4 x y + 6 y^{2} - 8 x - 10

f (x, y) = x e^{- 2 x^{2} - 2 y^{2}}

4 x y

f (x) = \frac{- 2 x ^{2} + 5 x - 1}{2 x - 1}

f (x) = 100 - x^{2} - y^{2}

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