# Tutor profile: Paul N.

## Questions

### Subject: Pre-Calculus

How do I find $$f\circ g(x)$$ where $$f(x)=x^2-2x-3$$ and $$g(x)=x-1$$

This is called a composition of two functions. $$f\circ g(x)=f(g(x))$$ For $$f(x)$$, $$x$$ is the independent variable. Here we want to replace the "x" in the function $$f(x)$$ by the second function $$g(x)$$. One helpful way is to visualize the first [outer] function with an empty space in place of the "x". Use parentheses around this empty space. Then we write the entire function $$g(x)$$ in this empty space. $$f(x)=x^2-2x-3$$ becomes $$f(--)=(--)^2-2(--)-3$$ Wherever you see the empty dashes, write in the expression: $$x-1$$ $$f(x-1)=(x-1)^2-2(x-1)-3$$ Use algebra to simplify. $$x^2-2x+1$$ $$-2x+2$$ $$-3$$ = $$x^2-4x$$ So $$f\circ g(x)= x^2-4x$$

### Subject: Trigonometry

How do I solve the equation $$sin(x)=1/2$$

This is a trigonometric equation. Since the sine function is periodic, there are infinitely many solutions to this. Sometimes a domain is included in the problem. The value is 1/2. We know this corresponds to a special angle, namely $$\frac{\pi}{6}$$ We also know that there are two angles in the unit circle whose sine is 1/2. Since 1/2 is positive, the angles have to lie in the first and second quadrants. The second quadrant angle is $$\pi-\frac{\pi}{6}=\frac{5\pi}{6}$$ The two solutions in the unit circle are: $$\frac{\pi}{6}$$ and $$\frac{5\pi}{6}$$ Since there were no restrictions on the domain, then we will have to show all of the solutions to the equation. The period of the sine function is $$2\pi$$ . The solutions will "repeat" every $$2n\pi$$ units, where n is an integer. Therefore the solutions to the equation are: $$\frac{\pi}{6}+2n\pi$$ and $$\frac{5\pi}{6}+2n\pi$$

### Subject: Algebra

How do I simplify the expression $$\frac{1}{x}-\frac{2}{x+1}$$

This is called a "rational expression". Like fractions, it is one of those topics that cause students some trouble. If you remember to follow the rules, you can conquer problems like this. {Please note that this is not a rational "equation" which has a different procedure.} Think of these two terms as fractions. When you add or subtract two fractions, you must have a "common denominator". In this case, we call this the LCD (least common denominator). Since the two denominators do not have any common factors, we will say that the LCD is the expression (x)(x+1) We need to rewrite each of the two separate terms with the same [matching] denominators. We accomplish this by multiplying the "top"[numerator] and "bottom"[denominator] of each fraction by the same expression. $$\frac{1}{x}\frac{(x+1)}{(x+1)}-\frac{2}{x+1}\frac{(x)}{(x)}$$ Notice that the denominators now match. Simplifying the expression, we get $$\frac{x+1}{(x)(x+1)}-\frac{2x}{(x)(x+1)}$$ Now that we have the same denominator, we combine (add or subtract) the numerators. Remember to keep the same denominator. $$\frac{x+1 - 2x}{(x)(x+1)}$$ Simplify the numerator $$\frac{1-x}{(x)(x+1)}$$ OR $$\frac{-x+1}{(x)(x+1)}$$ Please notice that there are no common factors in the numerator or denominator, so the expression cannot be reduced any further.

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