What is the extreme f(x)=x^4-5x^3+1.1x^2-10.2x+1 ?

The extreme f(x)=x^4-5x^3+1.1x^2-10.2x+1 is Minimum(3.78280…,-87.73138…)

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extreme f(x)=x^4-5x^3+1.1x^2-10.2x+1

extreme points

f (x) = x^{4} - 5 x^{3} + 1.1 x^{2} - 10.2 x + 1

Minimum (3.78280 \dots, - 87.73138 \dots)

Solution steps

Find the critical points:

x \approx 3.78280 \dots

Domain of

x^{4} - 5 x^{3} + 1.1 x^{2} - 10.2 x + 1 : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < 3.78280 \dots,

Increasing

: 3.78280 \dots < x < \infty

Plug the extreme point

x = 3.78280 \dots

into

x^{4} - 5 x^{3} + 1.1 x^{2} - 10.2 x + 1 \Rightarrow y = - 87.73138 \dots

Minimum (3.78280 \dots, - 87.73138 \dots)

(x - 1) (x + 9)^{2}

f (x) = 2 csc (x)

f (x) = (x^{2} - 36)^{\frac{1}{11}}, - 7 \leq x \leq 6

x^{\frac{1}{5}} (x^{2} - 4)

f (x) = cos (\frac{4 x}{3})

What is the extreme f(x)=x^4-5x^3+1.1x^2-10.2x+1 ?
The extreme f(x)=x^4-5x^3+1.1x^2-10.2x+1 is Minimum(3.78280…,-87.73138…)