What is the range of the function ?

Question

$y = \frac{4 x - 3}{2 x}$ $y=(4x-3)/(2x)$

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Answer 1 · 2017-12-28T00:16:41

$(- \infty, 2) \cup (2, \infty)$ $(-oo, 2) uu (2, oo)$

Given:

$y = \frac{4 x - 3}{2 x} = 2 - \frac{3}{2 x}$ $y = (4x-3)/(2x)= 2-3/(2x)$

Then:

$\frac{3}{2 x} = 2 - y$ $3/(2x)=2-y$

So taking the reciprocal of both sides:

$\frac{2}{3} x = \frac{1}{2 - y}$ $2/3x = 1/(2-y)$

Multiplying both sides by $\frac{3}{2}$ $3/2$ , this becomes:

$x = \frac{3}{2 (2 - y)}$ $x = 3/(2(2-y))$

So for any $y$ $y$ apart from $2$ $2$ , we can substitute $y$ $y$ into this formula to give us a value of $x$ $x$ that satisfies:

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

So the range is the whole of the real numbers except $2$ $2$ , i.e. it is:

$(- \infty, 2) \cup (2, \infty)$ $(-oo, 2) uu (2, oo)$

graph{y = (4x-3)/(2x) [-10, 10, -5, 5]}