A silly question on range of a quadratic function.

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

• 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.
– lulu
Jul 1 at 13:44
• You have shown that $x^2 + 3x + 2-y=0$ has a solution if and only if $y \ge -1/4$. Jul 1 at 13:46
• 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. Jul 2 at 9:08

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.

• That is certainly a better way to do it, but I was experimenting with the other ways there are to approach the problem. Jul 2 at 6:24
• @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 Jul 2 at 7:11

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. :)

• That makes sense, thank you! Jul 2 at 6:25

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}

• Thank you! This clears my doubt Jul 2 at 6:28

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.

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