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I want to find the range of: $$y=\sqrt {x^2+2x+3}$$

I would like to know if we can solve this by writing $x$ in terms of $y$ and then finding the domain of that? If so how?

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  • $\begingroup$ Range in real numbers for square root requires the expression to be greater or equal to zero. $\endgroup$
    – NoChance
    Oct 20, 2019 at 14:36
  • $\begingroup$ Where is the square root negative? When $x^2+2x+3<0$....there is the restriction for the inputs. Which then helps with the restrictions for the output... $\endgroup$ Oct 20, 2019 at 14:36

5 Answers 5

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Note

$$ y = \sqrt{ (x+1)^2 +2}\ge \sqrt2$$

for all real $x$.

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Since $B^2 < 4AC$ and $A>0$ so the quadratic is positive definite for all real values of $x$ so the domain of the funtion is $(-\infty, \infty)$ and the $f(x)=\sqrt{(x+1)^2+2}$ so the range is $[\sqrt{2}, \infty)$.

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It’s clear that as $x$ approaches infinity, the function approaches infinity. So we have to find the minimum of the quadratic.

The axis of symmetry is $\frac{-b}{2a}$ which is equal to $-1$. Then plug in $-$ into your quadratic and get $2$. So the minimum of your function is $2^{0.5}$

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Assuming $x \in \mathbb{R}$, we have $x^2 + 2x + 3 \geq 2$ for all $x \in \mathbb{R}$.

$\implies $ $y \geq \sqrt{2} $.

The range is $[\sqrt{2}, +\infty)$.

Or you can start by writing $x^2 + 2x + 3 = y^2$

$\implies $ $(x+ 1)^2 + 2 - y^2 = 0$ $\implies $ $x = -1 \pm \sqrt{y^2 -2}$.

Let $f(y) = -1 \pm \sqrt{y^2 - 2}$.

The set of values of $y$ for which $f(y) \in \mathbb{R}$ is $y^2 \geq 2$ $\implies $ $y \geq \sqrt{2}$.

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  • $\begingroup$ why >= 2? Why not >=0? $\endgroup$
    – NoChance
    Oct 20, 2019 at 14:37
  • $\begingroup$ But I want to know if we can write x in terms of y and then finding the domain of that? $\endgroup$ Oct 20, 2019 at 14:43
  • $\begingroup$ But did you write y≥ √2? $\endgroup$ Oct 20, 2019 at 14:56
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$y = \sqrt {x^2+2x+3} \text{ iff } y^2 = x^2+2x+3 \land y \ge 0 \text{ iff }$

$\quad x^2+2x+3 - y^2 = 0 \land y \ge 0 \text{ iff } (x+1)^2 + 2-y^2 = 0 \land y \ge 0 \text{ iff } $

$\quad (x+1)^2 = y^2 - 2 \land y \ge \sqrt 2 \text{ iff } x + 1 = \pm(\sqrt{y^2-2}) \land y \ge \sqrt 2 \text{ iff }$

$\quad y \ge \sqrt 2 \land \big[ x = \sqrt{y^2-2} - 1 \text{ or } x = -\sqrt{y^2-2} - 1 \big]$

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  • $\begingroup$ What's the vertical symbol you used "^".what does that mean? $\endgroup$ Oct 20, 2019 at 16:13
  • $\begingroup$ Logical symbol for conjunction $\quad \land = \text{and}$ $\endgroup$ Oct 20, 2019 at 16:18

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