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For example, if I evaluate $(-2)^2$ in my calculator, I get 4. However, if I evaluate $-2^2$ I get -4. I noticed this when doing my homework on functions, relating to this question:

Find the range of the function $y = x^2 - 1$ in the restricted domain of $-2 <= x <= 3$. Can someone else please show working out for this? I got + 8 part right, but the <= part is some how -1 !

astgeh
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  • Think of $-2^2$ as $0-2^2$. – David Mitra May 29 '14 at 12:09
  • Two different questions. The first one has been properly addressed. For the second one, you have to realize that the range $y$ when $x$ is restricted to $[-2, 3]$ is not $[y(-2), y(3)] = [3, 8]$. Think about it. The correct result is $[-1, 8]$ because when $x=0$, $y=-1$ and that's the lowest value $y$ can take. – Ant May 29 '14 at 12:19

3 Answers3

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Since exponents are prioritized in the order of operations (before multiplication) we have that $-2^2 = -(2^2) = -(2\cdot2) = -4.$ Likewise,$(-2)^2 = (-2)\cdot(-2) = 4$.

For the other question, i.e. find the range of $y = x^2-1$ with $-2\leq x\leq 3$, you have to think "when is this function at its minimum and when is it at its maximum?" As you saw $(-2)^2 = 4$, and in general we can say that $x^2\geq 0$ for all real numbers. So the minimum of $x^2$ is when $x=0$ and then $y=x^2-1 = 0^2-1 = -1$.

Eff
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Hint: $$(-2)^2=(-2)(-2)=4$$ $$-2^2=-(2^2)=-(2)(2)=-4$$

Rocket Man
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Exponentiation (E) occurs first, then multiplication (M), unless brackets (B) indicate otherwise.

We can consider $-2 = -1\times 2$.

So $-2^2 = -1\cdot \underbrace{2^2}_{\text{E before M}} = -1\cdot 4 = -4$.

On the other hand $\underbrace{(-1\cdot 2)^2}_{\text{B before E}} = (-2)^2 = -2\cdot -2 = 4$.

amWhy
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