What is the extreme f(x)=(x^4+1)/(x^2) ?

The extreme f(x)=(x^4+1)/(x^2) is Minimum(-1,2),Minimum(1,2)

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extreme f(x)=(x^4+1)/(x^2)

extreme points

f (x) = \frac{x ^{4} + 1}{x ^{2}}

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

Solution steps

f^{'} (x) = \frac{2 ( x ^{4} - 1 )}{x ^{3}}

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

\frac{x ^{4} + 1}{x ^{2}} : 2

Minimum (- 1, 2)

Plug

x = 1

into

\frac{x ^{4} + 1}{x ^{2}} : 2

Minimum (1, 2)

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

f (x, y) = x^{3} - 3 x y + y^{3}

x^{2} + y^{2}

f (x, y) = 4 x^{2} + 9 y^{2}

f (x) = 6 x^{5} - 40 x^{3} + 14

f (x) = - x^{4} + 4 x^{3} + 20

What is the extreme f(x)=(x^4+1)/(x^2) ?
The extreme f(x)=(x^4+1)/(x^2) is Minimum(-1,2),Minimum(1,2)