What is the range of f(x)=sqrt((x^2-4)/(x-2)) ?

The range of f(x)=sqrt((x^2-4)/(x-2)) is 0 2

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range of f(x)=sqrt((x^2-4)/(x-2))

range

f (x) = \frac{x ^{2} - 4}{x - 2}

0 \leq f (x) < 2 or f (x) > 2

Interval Notation

[0, 2) \cup (2, \infty)

Solution steps

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

Domain of

\frac{x ^{2} - 4}{x - 2} : - 2 \leq x < 2 or x > 2

Simplify

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

Extreme Points of

x + 2 :

Minimum

(- 2, 0)

Find the range for the interval

- 2 \leq x < 2 : 0 \leq f (x) < 2

Find the range for the interval

2 < x < \infty : 2 < f (x) < \infty

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

0 \leq f (x) < 2 or f (x) > 2

0 \leq f (x) < 2 or f (x) > 2

g (x) = 2 x + 1

f (x) = x^{45}

f (x) = \frac{7}{10} x - 13

f (x) = \frac{( 3 + x ^{4} )}{x ^{2} - x ^{4}}

f (x) = 3 x^{3} - 36 x - 8

What is the range of f(x)=sqrt((x^2-4)/(x-2)) ?
The range of f(x)=sqrt((x^2-4)/(x-2)) is 0<= f(x)<2\lor f(x)>2