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I'm asked to find the range of the function :

$ (-x^2 , +1) $

Taking R as a real number which represents a quantity along a line and graphing :

enter image description here

then the range for this function from viewing the graph appears to be :

$ R = (-\infty , +\infty) $

Is there an alternative method of finding the range of this function without using a graph ?

Watching the khan academy tutorial suggests using graphs : https://www.khanacademy.org/math/algebra/algebra-functions/domain-and-range/v/domain-and-range-from-graphs but is there a pure algebraic method instead of using graphs ?

It is not clear what the range is when the range appears infinite as how do we know that at some point on the axis the range functions stops tending towards infinity ?

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Notice that $x^2 \ge 0$ for all $x \in \Bbb R$, which means that $1 - x^2 \le 1$ for all $x \in \Bbb R$, with equality for $x=0$. Notice, too, that $x^2$ increases towards $\infty$, therefore $1-x^2$ will decrease towards $-\infty$. Finally, notice that $x^2$ is surjective (for every $y \ge 0$, there exist $\sqrt y$). Therefore, the values of $f$ lay between $- \infty$ and $1$, meaning that the range is $(-\infty, 1]$.

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  • $\begingroup$ Small addendum, because that is a bit unclear in your answer: $x^2$ is surjective as a function $\mathbb{R}\to[0,\infty)$, but not as a function $\mathbb{R}\to\mathbb{R}$. $\endgroup$ – Janik Sep 24 '16 at 10:25
  • $\begingroup$ @Janik: I've written "for every $\sqrt y \ge 0$", so this should say it all. $\endgroup$ – Alex M. Sep 24 '16 at 10:35
  • $\begingroup$ @AlexM. $1-x^2=0$ , can you elaborate how you arrive at $1-x^2<=1$ from this assertion ? $\endgroup$ – blue-sky Sep 24 '16 at 10:50
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    $\begingroup$ @blue-sky: I don't get your question, you must have misunderstood me. I'm saying that $x^2 \ge 0 \iff x^2 - 1 \ge -1 \iff -x^2 + 1 \le 1$ (for the last inequality multiply the preceding one by $-1$). $\endgroup$ – Alex M. Sep 24 '16 at 11:00
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You see that if we put x as infinity we get f(x)=-∞. Now to find the maximum value of functions just differentiate it and put the derivative equal to zero to find the point where the function stops increasing and starts decreasing. $\frac{d}{dx}(-x^2 + 1) = -2x$ $-2x=0 =>$ f(x) is maximum when x=0 Therefore maximum value of f(x) is +1 $f(x)\epsilon (-∞,+1) $ here is the max value on graph

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