What is the range of (x^3+5)/(sqrt(x)) ?

The range of (x^3+5)/(sqrt(x)) is f(x)>= 6

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range of (x^3+5)/(sqrt(x))

range

\frac{x ^{3} + 5}{x}

f (x) \geq 6

Interval Notation

[6, \infty)

Solution steps

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

Domain of

\frac{x ^{3} + 5}{x} : x > 0

Extreme Points of

\frac{x ^{3} + 5}{x} :

Minimum

(1, 6)

Find the range for the interval

0 < x < \infty : 6 \leq f (x) < \infty

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

f (x) \geq 6

(1, 2), (0, 5)

f (x) = sin (arcsin (x))

\frac{x + 3}{4}

\frac{x + 6}{x ^{2} + 10 x}

4 - x + x^{2} - 9

What is the range of (x^3+5)/(sqrt(x)) ?
The range of (x^3+5)/(sqrt(x)) is f(x)>= 6