What is the range of \sqrt[4]{x^2-6x} ?

The range of \sqrt[4]{x^2-6x} is f(x)>= 0

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range of \sqrt[4]{x^2-6x}

range

f (x) \geq 0

Interval Notation

[0, \infty)

Solution steps

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

Domain of

Extreme Points of

Minimum

(0, 0),

Minimum

(6, 0)

Find the range for the interval

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

Find the range for the interval

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

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

f (x) \geq 0 or f (x) \geq 0

f (x) \geq 0

x + 4 + 7

\frac{- 2 x + 4}{x ^{2} - 4}

\frac{x ^{3} + 1}{x ^{2}}

f (x) = \frac{1}{( x - 8 ) ^{2}}

What is the range of \sqrt[4]{x^2-6x} ?
The range of \sqrt[4]{x^2-6x} is f(x)>= 0