What is the extreme f(x)=(x^2+2x-7)/(x^2+2x-3) ?

The extreme f(x)=(x^2+2x-7)/(x^2+2x-3) is Minimum(-1,2)

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extreme f(x)=(x^2+2x-7)/(x^2+2x-3)

extreme points

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

Minimum (- 1, 2)

Solution steps

f^{'} (x) = \frac{8 x + 8}{( x ^{2} + 2 x - 3 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < - 1,

Increasing

: - 1 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

\frac{x ^{2} + 2 x - 7}{x ^{2} + 2 x - 3} : 2

Minimum (- 1, 2)

f (x) = 1 + x - 3 x^{2}

f (x) = \frac{x ^{2}}{2 x + 4}

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

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

f (x) = \frac{x}{x ^{2} - x + 1}, 0 \leq x \leq 3

What is the extreme f(x)=(x^2+2x-7)/(x^2+2x-3) ?
The extreme f(x)=(x^2+2x-7)/(x^2+2x-3) is Minimum(-1,2)