What is the extreme f(x)=4x-x^2[1.5] ?

The extreme f(x)=4x-x^2[1.5] is Maximum(4/3 ,2.66666…)

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extreme f(x)=4x-x^2[1.5]

extreme points

f (x) = 4 x - x^{2} [1.5]

Maximum (\frac{4}{3}, 2.66666 \dots)

Solution steps

f^{'} (x) = 4 - 3 x

Find intervals:

Increasing

: - \infty < x < \frac{4}{3},

Decreasing

: \frac{4}{3} < x < \infty

Plug

x = \frac{4}{3}

into

4 x - x^{2} [1.5] : 2.66666 \dots

Maximum (\frac{4}{3}, 2.66666 \dots)

f (x) = 9 x y e^{- x^{2} - y^{2}}

f (x) = x^{3} - 12 x - 1

f (x, y) = 9 - x^{2} - 9 y^{2}

6 sin (x)

2 t^{3} - 24 t^{2} + 70

What is the extreme f(x)=4x-x^2[1.5] ?
The extreme f(x)=4x-x^2[1.5] is Maximum(4/3 ,2.66666…)