What is the extreme f(x)= 1/2 (7x-2),x<= 3 ?

The extreme f(x)= 1/2 (7x-2),x<= 3 is Maximum(3, 19/2)

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extreme f(x)= 1/2 (7x-2),x<= 3

extreme points

f (x) = \frac{1}{2} (7 x - 2), x \leq 3

Maximum (3, \frac{19}{2})

Solution steps

Find the critical points:

x = 3

Domain of

\frac{1}{2} (7 x - 2), x \leq 3 : x \leq 3

Find intervals:

Increasing

: - \infty < x < 3

Plug the extreme point

x = 3

into

\frac{1}{2} (7 x - 2) \Rightarrow y = \frac{19}{2}

Maximum (3, \frac{19}{2})

\frac{1}{t ^{3} + 2}

f (x) = \frac{6}{7} (x^{2} - 9)^{\frac{2}{3}}

f (x, y) = 7 x y + 14 x - x^{2} + 2 y^{2}

4 x^{2} - 8 x^{4}

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

What is the extreme f(x)= 1/2 (7x-2),x<= 3 ?
The extreme f(x)= 1/2 (7x-2),x<= 3 is Maximum(3, 19/2)