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Let $f:\mathbb{R}\to\mathbb{R}\space|\space f(x) = x^2 + 3x + 2 \space\forall\space x \in\mathbb{R}$

We are asked to find the range of the function $f$

I begin as follows:

Assume $y=x^2 + 3x + 2$ for some $ x\in\mathbb{R}$

Collecting terms to one side,

$x^2 + 3x + 2-y=0$

Now as x is real, the discriminant of the above quadratic expression must be greater than or equal to zero. Hence,

$3^2 - 4(1)(2-y)\geqslant0$

Solving, we get,

$y\geqslant-1/4$

Hence the range of the function $f$ is the set $A=\{x|x\geq-1/4\}$

This answer is correct. The range of the function $f$ is indeed the set $A=\{x|x\geq-1/4\}$.

However, I have a problem with my last step. The final equality only tells me that $y$ is greater than OR equal to $-1/4$. It doesn't exactly say that $y$ WILL take all values greater than AND equal to $-1/4$. Is there a way to prove that $y$ will take all such values?

Any hint or help regarding this will be appreciated. Thank you

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    $\begingroup$ The function is continuous and approaches $+ \infty$ as $x\to \pm \infty$. It follows that, if $f(x)$ takes the value $y$, then it takes all greater values. $\endgroup$
    – lulu
    Jul 1 at 13:44
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    $\begingroup$ You have shown that $x^2 + 3x + 2-y=0$ has a solution if and only if $y \ge -1/4$. $\endgroup$
    – Martin R
    Jul 1 at 13:46
  • $\begingroup$ I think you're misinterpreting the symbol $\ge$. The set $\{y \in \mathbb{R} \mid y \ge -1/4\}$ includes the value $-1/4$. Each value may only be greater than or equal, but the solution includes both the values that are greater than, and the one value that is equal. $\endgroup$ Jul 2 at 9:08

4 Answers 4

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I would write this as $y = (x+\frac32)^2-\frac14$, from which it is obvious what the range would be, as the square term takes all nonnegative values.

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  • $\begingroup$ That is certainly a better way to do it, but I was experimenting with the other ways there are to approach the problem. $\endgroup$
    – neofyt
    Jul 2 at 6:24
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    $\begingroup$ @RajShukla Your approach works well too, as others noted you have proved for every $y\ge -\frac14$, there is at least one real value of $x$, and not for any other $y$, so the range is established $\endgroup$
    – Macavity
    Jul 2 at 7:11
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In your solution, you show that $x$ has a real value if $y\geq -1/4$ using discriminant of a quadratic. Well this works the other way too! If $y\geq -1/4$, then the quadratic has nonnegative discriminant and so you will always find $x$ that give that specific value of $y$.

Hope this helps. :)

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  • $\begingroup$ That makes sense, thank you! $\endgroup$
    – neofyt
    Jul 2 at 6:25
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Your calculation is correct and complete if you write it like this: $$ \begin{align} &\text{$y$ is in the range of $f$} \\ \iff &x^2 + 3x + 2-y=0 \text{ for some } x \in \Bbb R \\ \iff &\text{the discriminant of $x^2 + 3x + 2-y$ is non-negative} \\ \iff &y \ge -1/4 \, . \end{align} $$

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  • $\begingroup$ Thank you! This clears my doubt $\endgroup$
    – neofyt
    Jul 2 at 6:28
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I think the OP’s doubt is that “f(x)$\geq\frac 14$“ states that any values taken by f(x) are greater than or equal to $\frac 14$, but not that f(x) takes all values greater than or equal to $\frac 14$.


Note that $y=f(x)$ is continuous and differentiable on all points in $\mathbb R$.

The minimum value of a quadratic function $$ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a^2}$$ occurs at $\displaystyle x=-\frac{b}{2a}$ so there exists some x such that the minimum occurs.

Now, take an interval $\left[-\frac{b}{2a}, q\right]$ where $q\gt-\frac{b}{2a} $ is some real number. Then, by the Intermediate Value Theorem, the function is guaranteed to take on all the values between $\displaystyle -\frac{b}{2a}$ and $f(q)$. Notice that f(x) is increasing in $\left(-\frac{b}{2a},\infty\right)$ so $\displaystyle f(q)> -\frac{b}{2a}$. You can keep increasing q indefinitely and so we can conclude that f(x) assumes all values greater than and including $-\frac{b}{2a}$.
(P.S. a similar exercise can be carried out using $q’< -\frac{b}{2a}$.)

Thus, we have proved that the function is guaranteed to take on all the values greater than and including its minimum.

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  • $\begingroup$ Thank you! That clears a lot of stuff for me. $\endgroup$
    – neofyt
    Jul 2 at 6:26
  • $\begingroup$ Oh, yes. I’ll fix it. $\endgroup$ Jul 2 at 10:30

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