What is the extreme f(x)=x(20-2x)(32-2x),0<x<10 ?

The extreme f(x)=x(20-2x)(32-2x),0<x<10 is Maximum(4,1152)

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extreme f(x)=x(20-2x)(32-2x),0<x<10

extreme points

f (x) = x (20 - 2 x) (32 - 2 x), 0 < x < 10

Maximum (4, 1152)

Solution steps

Find the critical points:

x = 4

Domain of

x (20 - 2 x) (32 - 2 x), 0 < x < 10 : 0 < x < 10

Find intervals:

Increasing

: 0 < x < 4,

Decreasing

: 4 < x < 10

Plug the extreme point

x = 4

into

x (20 - 2 x) (32 - 2 x) \Rightarrow y = 1152

Maximum (4, 1152)

f (x) = 3 ∣ x - 1 ∣, 0 < x < 3

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

f (x, y) = x^{3} + y^{3} + x y

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

x^{2} - 4 x + 1

What is the extreme f(x)=x(20-2x)(32-2x),0<x<10 ?
The extreme f(x)=x(20-2x)(32-2x),0<x<10 is Maximum(4,1152)