What is the range of f(x)=sqrt(1/3 (x-1)) ?

The range of f(x)=sqrt(1/3 (x-1)) is f(x)>= 0

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range of f(x)=sqrt(1/3 (x-1))

range

f (x) = \frac{1}{3} (x - 1)

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

\frac{1}{3} (x - 1) : x \geq 1

Extreme Points of

\frac{1}{3} (x - 1) :

Minimum

(1, 0)

Find the range for the interval

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

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

f (x) \geq 0

f (x) = y^{2} = - 16 x

f (x) = \frac{x}{x ^{2} + x - 6}

f (x) = \frac{( e ^{x} )}{8 + e ^{x}}

f (x) = 3 x^{4} - 24 x^{2} + 18

y - 11 = 0 (x + 3)

What is the range of f(x)=sqrt(1/3 (x-1)) ?
The range of f(x)=sqrt(1/3 (x-1)) is f(x)>= 0