What is the range of (2sqrt(x))/(x^2+6) ?

The range of (2sqrt(x))/(x^2+6) is 0<= f(x)<= 1/(2*2^{3/4)}

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range of (2sqrt(x))/(x^2+6)

range

\frac{2 x}{x ^{2} + 6}

0 \leq f (x) \leq \frac{1}{2 \cdot 2 ^{\frac{3}{4}}}

Interval Notation

[0, \frac{1}{2 2 ^{\frac{3}{4}}}]

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

\frac{2 x}{x ^{2} + 6} : x \geq 0

Extreme Points of

\frac{2 x}{x ^{2} + 6} :

Minimum

(0, 0),

Maximum

(2, \frac{1}{2 \cdot 2 ^{\frac{3}{4}}})

Find the range for the interval

0 \leq x < \infty : 0 \leq f (x) \leq \frac{1}{2 \cdot 2 ^{\frac{3}{4}}}

Combine the ranges of all domain intervals to obtain the function range

0 \leq f (x) \leq \frac{1}{2 \cdot 2 ^{\frac{3}{4}}}

f (x) = \frac{3 ( x + 10 )}{5}

f (x) = e^{- 2 x}

4 x - 3 y = - 21

- 6

f (x) = 3 - 2 x

What is the range of (2sqrt(x))/(x^2+6) ?
The range of (2sqrt(x))/(x^2+6) is 0<= f(x)<= 1/(2*2^{3/4)}