What is the extreme f(x)=((x^2+x+4))/((3+x^2)) ?

The extreme f(x)=((x^2+x+4))/((3+x^2)) is Minimum(-3, 5/6),Maximum(1, 3/2)

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extreme f(x)=((x^2+x+4))/((3+x^2))

extreme points

f (x) = \frac{( x ^{2} + x + 4 )}{( 3 + x ^{2} )}

Minimum (- 3, \frac{5}{6}), Maximum (1, \frac{3}{2})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 3

into

\frac{x ^{2} + x + 4}{3 + x ^{2}} : \frac{5}{6}

Minimum (- 3, \frac{5}{6})

Plug

x = 1

into

\frac{x ^{2} + x + 4}{3 + x ^{2}} : \frac{3}{2}

Maximum (1, \frac{3}{2})

Minimum (- 3, \frac{5}{6}), Maximum (1, \frac{3}{2})

(1 - x^{2})^{2}

f (x) = 3 + 9 x - 3 x^{2} - x^{3}

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

f (x) = - \frac{10}{3} x^{3} - 19 x^{2} - 24 x + 50

y = 4 cos (\frac{x}{2}) + 2

What is the extreme f(x)=((x^2+x+4))/((3+x^2)) ?
The extreme f(x)=((x^2+x+4))/((3+x^2)) is Minimum(-3, 5/6),Maximum(1, 3/2)