Question

Find the range of the real valued function: f(x) = $\frac{x^{2} - 4} {x - 2}$

Answer 1

Answer:

Domain = R - {2}

Range = R - {0}

The given function is

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

It is not valid for

x - 2 = 0

⇒ x = 2

So  x ≠ 2

Domain = R - {2}

                  And  as   x→2  ,  f(x)→0

                  But x ≠ 2 , so f(x) ≠ 0

Range = R - {0}

Answer 2

Answer:

range(f)=R

Step-by-step explanation:

To find the range of real valued function

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

it is clear that domain of f(x) does not include 2

dom(f)=R-{2}

So range

$y=\frac{x^{2}-4}{x-2} \\\\y(x-2)=(x-2)(x+2)\\\\x=y-2$

here now check for the value of y for which the function has finite value.

here for all values of y f(y) exist.

so range(f)=R