What is the f(x)=(4x)/(x^2+9),-4<= x<= 4 ?

The f(x)=(4x)/(x^2+9),-4<= x<= 4 is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

f(x)=(4x)/(x^2+9),-4<= x<= 4

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

Domain : - 4 \leq x \leq 4

Range : - \frac{2}{3} \leq f (x) \leq \frac{2}{3}

X Intercepts : (0, 0), Y Intercepts : (0, 0)

Asymptotes : Horizontal None

Extreme Points : Maximum (- 4, - \frac{16}{25}), Minimum (- 3, - \frac{2}{3}), Maximum (3, \frac{2}{3}), Minimum (4, \frac{16}{25})

Interval Notation

Domain : [- 4, 4]

Range : [- \frac{2}{3}, \frac{2}{3}]

Solution steps

Domain of

\frac{4 x}{x ^{2} + 9}, - 4 \leq x \leq 4 : - 4 \leq x \leq 4

Range of

\frac{4 x}{x ^{2} + 9}, - 4 \leq x \leq 4 : - \frac{2}{3} \leq f (x) \leq \frac{2}{3}

Axis interception points of

\frac{4 x}{x ^{2} + 9}, - 4 \leq x \leq 4 :

X Intercepts

: (0, 0),

Y Intercepts

: (0, 0)

Asymptotes of

\frac{4 x}{x ^{2} + 9}, - 4 \leq x \leq 4 :

Horizontal

:

None

Extreme Points of

\frac{4 x}{x ^{2} + 9}, - 4 \leq x \leq 4 :

Maximum

(- 4, - \frac{16}{25}),

Minimum

(- 3, - \frac{2}{3}),

Maximum

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

Minimum

(4, \frac{16}{25})

y = ((9 x^{2}) (\frac{2}{6} x))

f (x) = 2 + lo g_{2} (x - 3)

f (x) = 2 x^{3} cos (2 x^{4})

f (x) = x^{4} - x - 1

f (x) = \frac{5}{x ^{2} + 4}