What's wrong in my solution and pls provide the correct one in this question of finding range

Asked May 27 '23 at 16:47

Active May 27 '23 at 17:58

Viewed 58 times

Find the range of the function $$y=\frac{\sqrt{(x-1)(x+3)}}{x+2}$$

My attempt: $$y=\frac{\sqrt{(x-1)(x+3)}}{x+2}$$ $$\implies x^2(y^2-1)+x(4y^2-2)+4y^2+3=0$$ Since $x\in\mathbb{R}$ so roots can't be complex $\implies \Delta\ge0$ $$(4y^2-2)^2-4(y^2-1)(4y^2+3)\ge0$$ $$\implies \frac{-2}{\sqrt3}\le y\le\frac{2}{\sqrt3}$$

But this isn't the answer. What's wrong$?$

Any help is greatly appreciated.

edited May 27 '23 at 17:58

asked May 27 '23 at 16:47

MathStackexchangeIsNotSoBad

4,716

Are you assuming that$$y=\frac{\sqrt{(x-1)(x+3)}}{x-2}\iff y^2=\frac{(x-1)(x+3)}{(x-2)^2}?$$ – Another User May 27 '23 at 17:16
@AnotherUser yess – MathStackexchangeIsNotSoBad May 27 '23 at 17:19
As an indication that things have indeed gone wrong, suppose $x\approx 2$. Then the numerator is approximately $\sqrt{5}$ but the denominator is some small quantity. Hence their ratio can be made arbitrarily large, so the range will not be a finite interval. – Semiclassical May 27 '23 at 17:19
@Semiclassical range is a finite interval according to wolframalpha – MathStackexchangeIsNotSoBad May 27 '23 at 17:20
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Not according to what I plugged into WA (“range of sqrt((x-1)(x+3))/(x-2)”). – Semiclassical May 27 '23 at 17:29
Oh sorry @Semiclassical – MathStackexchangeIsNotSoBad May 27 '23 at 17:58
I've edited the question – MathStackexchangeIsNotSoBad May 27 '23 at 17:58
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You are assuming that $x$ can take any real value ; this is not the case : values of $x$ in $(-3,1)$ have to be excluded in order we can take the square root. – Jean Marie May 27 '23 at 18:47
By the unfortunately hyper-ambiguous word "range" of a function f, can we assume that you mean the set of all f(x) such that f(x) makes sense (as a real number)? (A better word for that is the image of f, notated im(f).) – Dan Asimov May 27 '23 at 19:20

What's wrong in my solution and pls provide the correct one in this question of finding range

0 Answers0