What is the extreme f(x)=7x^4-2x^2-8x+8 ?

The extreme f(x)=7x^4-2x^2-8x+8 is Minimum(0.73067…,3.08206…)

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extreme f(x)=7x^4-2x^2-8x+8

extreme points

f (x) = 7 x^{4} - 2 x^{2} - 8 x + 8

Minimum (0.73067 \dots, 3.08206 \dots)

Solution steps

Find the critical points:

x \approx 0.73067 \dots

Domain of

7 x^{4} - 2 x^{2} - 8 x + 8 : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < 0.73067 \dots,

Increasing

: 0.73067 \dots < x < \infty

Plug the extreme point

x = 0.73067 \dots

into

7 x^{4} - 2 x^{2} - 8 x + 8 \Rightarrow y = 3.08206 \dots

Minimum (0.73067 \dots, 3.08206 \dots)

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

f (x) = \frac{( x ^{2} - 63 )}{x - 8}

x^{3} + 8 y^{3} - 6 x y + 5

f (x) = 4 x^{3} - 60 x^{2} + 200 x

f (x) = x^{3} + 2 y^{2} + 3 x y + 8

What is the extreme f(x)=7x^4-2x^2-8x+8 ?
The extreme f(x)=7x^4-2x^2-8x+8 is Minimum(0.73067…,3.08206…)