What is the extreme f(x)=(4x+9)/(x^2+9) ?

The extreme f(x)=(4x+9)/(x^2+9) is Minimum(-6,-1/3),Maximum(3/2 , 4/3)

Popular >

extreme f(x)=(4x+9)/(x^2+9)

extreme points

f (x) = \frac{4 x + 9}{x ^{2} + 9}

Minimum (- 6, - \frac{1}{3}), Maximum (\frac{3}{2}, \frac{4}{3})

Solution steps

f^{'} (x) = \frac{- 4 x ^{2} - 18 x + 36}{( x ^{2} + 9 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 6,

Increasing

: - 6 < x < \frac{3}{2},

Decreasing

: \frac{3}{2} < x < \infty

Plug

x = - 6

into

\frac{4 x + 9}{x ^{2} + 9} : - \frac{1}{3}

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

Plug

x = \frac{3}{2}

into

\frac{4 x + 9}{x ^{2} + 9} : \frac{4}{3}

Maximum (\frac{3}{2}, \frac{4}{3})

Minimum (- 6, - \frac{1}{3}), Maximum (\frac{3}{2}, \frac{4}{3})

f (x, y) = 2 x y - 1

f (2.4) = 3 - 6 (x - 2) + 9 (y - 4)

- 4 x^{4} - 9 x^{3} + 2 x^{2} + 7 x - 2

f (x) = e^{- 4 x - 4 y}

\frac{x - 2}{x + 1}

What is the extreme f(x)=(4x+9)/(x^2+9) ?
The extreme f(x)=(4x+9)/(x^2+9) is Minimum(-6,-1/3),Maximum(3/2 , 4/3)