Find, without graphing, the range of the function. $$y=x^2-5, x∈[-2,0]$$
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?
 A: 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.
A: 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 $(-2, 0)$.   
A: 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 $[-5, -1]$.
A: 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:
$$\calc
    \langle \exists x :: y = x^2-5 \;\land\; x \in \left[{-2},0\right] \rangle
\op\equiv\hint{isolate $\;x\;$}
    \langle \exists x :: 0 \leq y+5 \;\land\; (x = {-\sqrt{y+5}} \;\lor\; x = \sqrt{y+5}) \;\land\; {-2} \leq x \leq 0 \rangle
\op\equiv\hint{substitute for $\;x\;$, twice}
    0 \leq y+5 \;\land\; ({-2} \leq {-\sqrt{y+5}} \leq 0 \;\lor\; {-2} \leq \sqrt{y+5} \leq 0)
\op\equiv\hint{arithmetic; simplify using $\;0 \leq \sqrt{\cdot}\;$}
    {-5} \leq y \;\land\; (\sqrt{y+5} \leq 2 \;\lor\; \sqrt{y+5} = 0)
\op\equiv\hint{LHS of $\;\lor\;$ implies RHS}
    {-5} \leq y \;\land\; \sqrt{y+5} \leq 2
\op\equiv\hint{arithmetic}
    {-5} \leq y \;\land\; y \leq {-1}
\endcalc$$
$%
\endgroup
%$
