What is the global 8x^4-8x^2+1 ?

The global 8x^4-8x^2+1 is Global Minimum(-1/(sqrt(2)),-1),Global Minimum(1/(sqrt(2)),-1)

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global 8x^4-8x^2+1

global extreme points

8 x^{4} - 8 x^{2} + 1

Global Minimum (- \frac{1}{2}, - 1), Global Minimum (\frac{1}{2}, - 1)

Solution steps

Find the extreme points:

Minimum

(- \frac{1}{2}, - 1),

Maximum

(0, 1),

Minimum

(\frac{1}{2}, - 1)

Find the function range:

[- 1, \infty)

Verify extreme points are global points

Global Minimum (- \frac{1}{2}, - 1), Global Minimum (\frac{1}{2}, - 1)

\frac{x ^{2} - 2 x + 1}{x ^{3} - 3 x ^{2}}

f (x) = 6 x + 4

\frac{4 x ^{2} + 6 x - 4}{2 x ^{2} + 13 x + 15}

f (x) = 2 + x - 3

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

What is the global 8x^4-8x^2+1 ?
The global 8x^4-8x^2+1 is Global Minimum(-1/(sqrt(2)),-1),Global Minimum(1/(sqrt(2)),-1)