What is the extreme f(x)=4.5x^2-7.1x+3.9 ?

The extreme f(x)=4.5x^2-7.1x+3.9 is Minimum(0.78888…,1.09944…)

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extreme f(x)=4.5x^2-7.1x+3.9

extreme points

f (x) = 4.5 x^{2} - 7.1 x + 3.9

Minimum (0.78888 \dots, 1.09944 \dots)

Solution steps

f^{'} (x) = 9 x - 7.1

Find intervals:

Decreasing

: - \infty < x < 0.78888 \dots,

Increasing

: 0.78888 \dots < x < \infty

Plug

x = 0.78888 \dots

into

4.5 x^{2} - 7.1 x + 3.9 : 1.09944 \dots

Minimum (0.78888 \dots, 1.09944 \dots)

y = 6 - 2 x^{2} - z^{2}

x^{3} - 9 x^{2} + 24 x - 12

f (x) = (1 - 5 x)^{3} x - 3

9 - x^{2}, - 3 \leq x \leq 2

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

What is the extreme f(x)=4.5x^2-7.1x+3.9 ?
The extreme f(x)=4.5x^2-7.1x+3.9 is Minimum(0.78888…,1.09944…)