What is the range of f(x)=(6-x)^{1/8} ?

The range of f(x)=(6-x)^{1/8} is f(x)>= 0

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range of f(x)=(6-x)^{1/8}

range

f (x) = (6 - x)^{\frac{1}{8}}

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

(6 - x)^{\frac{1}{8}} : x \leq 6

Extreme Points of

(6 - x)^{\frac{1}{8}} :

Minimum

(6, 0)

Find the range for the interval

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

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

f (x) \geq 0

f (x) = - (x + 1)^{2} + 4

sin (2 x)

x^{2} + 2 x

f (x) = x^{3} - 12

f (x) = \frac{1}{( x + 3 ) ^{2}}

What is the range of f(x)=(6-x)^{1/8} ?
The range of f(x)=(6-x)^{1/8} is f(x)>= 0