What is the extreme f(x)= x/(x^2-x+25),0<= x<= 15 ?

The extreme f(x)= x/(x^2-x+25),0<= x<= 15 is Minimum(0,0),Maximum(5, 1/9),Minimum(15, 3/47)

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extreme f(x)= x/(x^2-x+25),0<= x<= 15

extreme points

f (x) = \frac{x}{x ^{2} - x + 25}, 0 \leq x \leq 15

Minimum (0, 0), Maximum (5, \frac{1}{9}), Minimum (15, \frac{3}{47})

Solution steps

Find the critical points:

x = 0, x = 5, x = 15

Domain of

\frac{x}{x ^{2} - x + 25}, 0 \leq x \leq 15 : 0 \leq x \leq 15

Find intervals:

Increasing

: 0 < x < 5,

Decreasing

: 5 < x < 15

Plug the extreme point

x = 0

into

\frac{x}{x ^{2} - x + 25} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 5

into

\frac{x}{x ^{2} - x + 25} \Rightarrow y = \frac{1}{9}

Maximum (5, \frac{1}{9})

Plug the extreme point

x = 15

into

\frac{x}{x ^{2} - x + 25} \Rightarrow y = \frac{3}{47}

Minimum (15, \frac{3}{47})

Minimum (0, 0), Maximum (5, \frac{1}{9}), Minimum (15, \frac{3}{47})

f (x) = sin (x) + cos (x), (0, 2 π)

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

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

f (x) = 4 - x^{2}, - 2 \leq x \leq 1

x y

What is the extreme f(x)= x/(x^2-x+25),0<= x<= 15 ?
The extreme f(x)= x/(x^2-x+25),0<= x<= 15 is Minimum(0,0),Maximum(5, 1/9),Minimum(15, 3/47)