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Kevin Y.

High School Math Teacher for 13 years.

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Pre-Calculus

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Question:

What is the domain of the function $$f(x) = \frac{1}{\sqrt{x^2-1}}$$? State using interval notation.

Kevin Y.

Answer:

Domain can be restricted in several different ways. This problem shows two of the ways domain can be restricted. 1. Dividing by 0 2. Square root of a negative number To answer this question, we need to figure out what values of x make the denominator 0 and what values of x make the radicand (the number under the square root), negative. What makes the denominator equal to 0? To find that, set $${\sqrt{x^2-1}}$$ equal to $$0$$. $${\sqrt{x^2-1}}=0$$ $$x^2-1=0$$ $$x^2 = 1$$ $$x=1$$ and $$x=-1$$. To find what makes the radicand negative, we do the following: $$x^2-1>0$$ $$(x+1)(x-1)>0$$ This gives us $$x=-1$$ and $$x=1$$ as the boundaries with 3 test intervals. $$x<-1$$ $$-1<x<1$$ $$x>1$$ By testing values in each interval, I can see that $$x<-1$$ and $$x>1$$ produce true statements. Putting it all together, the domain can not include $$x=-1$$ or $$x=1$$ because that would create a 0 in the denominator. Also, $$x$$ can not be between $$-1$$ and $$1$$ because that would create a negative number under the radical. Therefore, the domain is $$(-\infty,-1)\cup(1,\infty)$$.

Pre-Algebra

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Question:

Why does anything to the zero power (except 0) equal one? For example, why is $$3^0 = 1$$?

Kevin Y.

Answer:

$$3^0$$ must be $$1$$ so that the rules of exponents remain consistent. For example, $$\frac{3^5}{3^2} = 3^{5-2} = 3^3$$ because $$\frac{3^5}{3^2} = \frac{3*3*3*3*3}{3*3} = 3*3*3 = 3^3$$ Therefore, when you divide two exponents that have the same base (in this case 3 is the base) you subtract the exponents. Returning to the original problem, let's write $$3^0$$ as a division problem. $$\frac{3^7}{3^7} = 3^{7-7} = 3^0$$ and, since anything divided by itself is 1, $$\frac{3^7}{3^7} = 1$$. Therefore, $$3^0 = 1$$.

Algebra

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Question:

If $$a^2 + b^2 = 4ab$$, what is $$(\frac{a+b}{a-b})^2$$?

Kevin Y.

Answer:

There are several ways to look at this problem that can highlight several skills covered in Algebra. First, consider $$(\frac{a+b}{a-b})^2$$. $$(\frac{a+b}{a-b})^2 = \frac{(a+b)^2}{(a-b)^2} = \frac{(a+b)(a+b)}{(a-b)(a-b)}$$ by the power of a quotient property. $$\frac{(a+b)(a+b)}{(a-b)(a-b)} = \frac{a^2+ab+ab+b^2}{a^2-ab-ab+b^2} = \frac{a^2+2ab+b^2}{a^2-2ab+b^2}$$ Now look back at the original problem and recall that $$a^2+b^2=4ab$$. Therefore, using the commutative property and substitution we can say $$\frac{a^2+2ab+b^2}{a^2-2ab+b^2} = \frac{a^2+b^2+2ab}{a^2+b^2-2ab} = \frac{4ab+2ab}{4ab-2ab} = \frac{6ab}{2ab} = 3$$ Cool. What if we consider $$a^2 + b^2 = 4ab$$ first? I could create two perfect square trinomials by adding or subtracting $$2ab$$ from each side. $$a^2 + b^2 = 4ab$$ $$a^2+2ab+b^2 = 4ab+2ab$$ $$(a+b)^2 = 6ab$$ $$a^2 + b^2 = 4ab$$ $$a^2-2ab+b^2 = 4ab-2ab$$ $$(a-b)^2 = 2ab$$ Therefore, $$(\frac{a+b}{a-b})^2 = \frac{(a+b)^2}{(a-b)^2}=\frac{6ab}{2ab}=3$$

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