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

The extreme f(x)=(x^2)/(x^2+2x-15) is Maximum(0,0),Minimum(15, 15/16)

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extreme f(x)=(x^2)/(x^2+2x-15)

extreme points

f (x) = \frac{x ^{2}}{x ^{2} + 2 x - 15}

Maximum (0, 0), Minimum (15, \frac{15}{16})

Solution steps

f^{'} (x) = \frac{2 x ^{2} - 30 x}{( x ^{2} + 2 x - 15 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 5,

Increasing

: - 5 < x < 0,

Decreasing

: 0 < x < 3,

Decreasing

: 3 < x < 15,

Increasing

: 15 < x < \infty

Plug

x = 0

into

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

Maximum (0, 0)

Plug

x = 15

into

\frac{x ^{2}}{x ^{2} + 2 x - 15} : \frac{15}{16}

Minimum (15, \frac{15}{16})

Maximum (0, 0), Minimum (15, \frac{15}{16})

f (x, y) = 2 x^{2} - 5 x y + 3 y^{4} + 5

f (x, y) = x^{3} + 3 x y^{2} - 3 x^{2} - 3 y^{2} + 4

f (x) = sin (π \cdot x)

- 3^{x + 1.5} + 2.94

f (x) = \frac{x ^{2} - 48}{x - 7}

What is the extreme f(x)=(x^2)/(x^2+2x-15) ?
The extreme f(x)=(x^2)/(x^2+2x-15) is Maximum(0,0),Minimum(15, 15/16)