What is the extreme f(x)=(2x+5)/3 ,0<= x<= 5 ?

The extreme f(x)=(2x+5)/3 ,0<= x<= 5 is Minimum(0, 5/3),Maximum(5,5)

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extreme f(x)=(2x+5)/3 ,0<= x<= 5

extreme points

f (x) = \frac{2 x + 5}{3}, 0 \leq x \leq 5

Minimum (0, \frac{5}{3}), Maximum (5, 5)

Solution steps

Find the critical points:

x = 0, x = 5

Domain of

\frac{2 x + 5}{3}, 0 \leq x \leq 5 : 0 \leq x \leq 5

Find intervals:

Increasing

: 0 < x < 5

Plug the extreme point

x = 0

into

\frac{2 x + 5}{3} \Rightarrow y = \frac{5}{3}

Minimum (0, \frac{5}{3})

Plug the extreme point

x = 5

into

\frac{2 x + 5}{3} \Rightarrow y = 5

Maximum (5, 5)

Minimum (0, \frac{5}{3}), Maximum (5, 5)

y = - 5 (x - 4)^{4} + 2

x^{2} + 4 x y + y^{2} - 40 x - 56 y + 1

x e^{3 - (\frac{x}{4})}

x y (1 - x - y)

f (x, y) = \frac{x ^{3} + 5 x y ^{2}}{y}

What is the extreme f(x)=(2x+5)/3 ,0<= x<= 5 ?
The extreme f(x)=(2x+5)/3 ,0<= x<= 5 is Minimum(0, 5/3),Maximum(5,5)