Find, without graphing, the range of the function.

Question

$$y=x^2-5, x∈[-2,0]$$ Here's what I did: $$-2≤x≤0$$ $$x^2≤4 ∧ x^2≤0$$ $$x^2≤0$$ $$x^2-5≤0-5$$ $$y≤-5$$ Is it correct?

Answer 1

If $0\leq a \leq b$, then $0\leq a^2 \leq b^2$. We use that as follows:

Since $-2\leq x \leq 0$, $0 \leq -x \leq 2$ so (applying the first line) $0 \leq (-x)^2 \leq 2^2$, i.e. $0 \leq x^2 \leq 4$. Apply minus five at both sides and you're done.

Answer 2

The function $y = x^2 - 5$ is decreasing on the interval $[-2, 0]$. As $x$ decreases from $-2$ to $0$, $x^2$ decreases from $4$ to $0$, so $x^2 - 5$ decreases from $4 - 5 = -1$ to $0 - 5 = -5$. Hence, the range is $[-5, -1]$.

Note: If you are familiar with calculus, you can demonstrate that $y$ is decreasing on the interval $[-2, 0]$ by showing that the derivative is negative at each point of the interval $(-2, 0)$.

Answer 3

The first and second derivatives of the function are $y' = 2x$ and $y'' = 2$, respectively. From this we can see that the only extremum over the reals is at $x=0$, and that it's a minimum.

Hence, the minimum of the range on the interval $[-2,0]$ is at $x=0$, so $y=-5$ is the minimum. The maximum is at the other end of the interval (as it increases monotonically as $x$ goes negative), so the maximum in the range is $y=(-2)^2 - 5 = -1$.

So, your range is $[-5, -1]$.

Answer 4

To answer this age-old OP-gone question, the following seems the most direct approach.$% \require{begingroup} \begingroup \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\Ref}[1]{\text{(#1)}} \newcommand{\then}{\Rightarrow} \newcommand{\when}{\Leftarrow} %$

Essentially we are asked a simplification question, that is, which $\;y\;$ satisfy $\Ref{0}$? Let's calculate:

$$\calc \langle \exists x :: y = x^2-5 \;\land\; x \in \left[{-2},0\right] \rangle \op\equiv\hint{isolate $\;x\;$} \langle \exists x :: 0 \leq y+5 \;\land\; (x = {-\sqrt{y+5}} \;\lor\; x = \sqrt{y+5}) \;\land\; {-2} \leq x \leq 0 \rangle \op\equiv\hint{substitute for $\;x\;$, twice} 0 \leq y+5 \;\land\; ({-2} \leq {-\sqrt{y+5}} \leq 0 \;\lor\; {-2} \leq \sqrt{y+5} \leq 0) \op\equiv\hint{arithmetic; simplify using $\;0 \leq \sqrt{\cdot}\;$} {-5} \leq y \;\land\; (\sqrt{y+5} \leq 2 \;\lor\; \sqrt{y+5} = 0) \op\equiv\hint{LHS of $\;\lor\;$ implies RHS} {-5} \leq y \;\land\; \sqrt{y+5} \leq 2 \op\equiv\hint{arithmetic} {-5} \leq y \;\land\; y \leq {-1} \endcalc$$

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