What is the extreme f(x)=x^4-3x^2+3x+1 ?

The extreme f(x)=x^4-3x^2+3x+1 is Minimum(-1.42366…,-5.24345…)

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extreme f(x)=x^4-3x^2+3x+1

extreme points

f (x) = x^{4} - 3 x^{2} + 3 x + 1

Minimum (- 1.42366 \dots, - 5.24345 \dots)

Solution steps

Find the critical points:

x \approx - 1.42366 \dots

Domain of

x^{4} - 3 x^{2} + 3 x + 1 : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < - 1.42366 \dots,

Increasing

: - 1.42366 \dots < x < \infty

Plug the extreme point

x = - 1.42366 \dots

into

x^{4} - 3 x^{2} + 3 x + 1 \Rightarrow y = - 5.24345 \dots

Minimum (- 1.42366 \dots, - 5.24345 \dots)

f (x) = \frac{x - 7}{x ^{2} + 5 x - 14}

f (x) = \frac{15}{1 + 4 e ^{- 0.2 x}}

\frac{1}{x + 6}

f (x) = \frac{3 x + 3}{5}

f (x) = ln (x - 2) + 1

What is the extreme f(x)=x^4-3x^2+3x+1 ?
The extreme f(x)=x^4-3x^2+3x+1 is Minimum(-1.42366…,-5.24345…)