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

The extreme f(x)=(x^2+2)/(x^2+3) is Minimum(0, 2/3)

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

extreme points

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

Minimum (0, \frac{2}{3})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

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

Minimum (0, \frac{2}{3})

- x^{2} - x + 1

\frac{x ^{2} + x}{∣ x ∣ + 1}

18 x y - 3 x^{2} y - 2 x y^{2}

f (x) = x^{2} - 8 x - 9, 1 \leq x \leq 6

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

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