What is the extreme f(x)= 1/2 (3x-1),x<= 3 ?

The extreme f(x)= 1/2 (3x-1),x<= 3 is Maximum(3,4)

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extreme f(x)= 1/2 (3x-1),x<= 3

extreme points

f (x) = \frac{1}{2} (3 x - 1), x \leq 3

Maximum (3, 4)

Solution steps

Find the critical points:

x = 3

Domain of

\frac{1}{2} (3 x - 1), x \leq 3 : x \leq 3

Find intervals:

Increasing

: - \infty < x < 3

Plug the extreme point

x = 3

into

\frac{1}{2} (3 x - 1) \Rightarrow y = 4

Maximum (3, 4)

f (x, y) = x^{3} - 12 x y + y^{3}

f (x) = 3 x ln (x)

f (x, y) = 5 x^{2} + 5 y^{2} + 5 x y - 10 x - 5 y + 18

f (x, y) = x^{3} + 3 x y^{2} - 15 x^{2} - 15 y^{2} + 72 x

f (x, y) = 10 - 4 x - 5 y

What is the extreme f(x)= 1/2 (3x-1),x<= 3 ?
The extreme f(x)= 1/2 (3x-1),x<= 3 is Maximum(3,4)