What is the range of f(x)=sqrt(x+3)*(8/x) ?

The range of f(x)=sqrt(x+3)*(8/x) is -infinity <f(x)<infinity

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range of f(x)=sqrt(x+3)*(8/x)

range

f (x) = x + 3 \cdot (\frac{8}{x})

- \infty < f (x) < \infty

Interval Notation

(- \infty, \infty)

Solution steps

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

Domain of

x + 3 \cdot \frac{8}{x} : - 3 \leq x < 0 or x > 0

Simplify

x + 3 \frac{8}{x} : \frac{8 x + 3}{x}

Extreme Points of

\frac{8 x + 3}{x} :

Maximum

(- 3, 0)

Find the range for the interval

- 3 \leq x < 0 : - \infty < f (x) \leq 0

Find the range for the interval

0 < x < \infty : 0 < f (x) < \infty

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

f (x) \leq 0 or f (x) > 0

- \infty < f (x) < \infty

f (x) = e^{2 x}

f (x) = - (0.4)^{x - 6}

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

f (x) = 8 + 9 x

- 4 x + 3

What is the range of f(x)=sqrt(x+3)*(8/x) ?
The range of f(x)=sqrt(x+3)*(8/x) is -infinity <f(x)<infinity