Section+3.6+10-27

math {}$ Today in class we started looking at Investigation 3B on Rational Functions. Our focus today was on Section 3.6: Reciprocal Functions. A lot of it was review, and for most of the class, we just looked at FYE #1-5 to review the issues we learned about last year. Here are the key ideas from each of the problems.

We observed what reciprocal functions are, one example being $g(x)=\frac{1}{f(x)}$. We also saw that if you have this same function, as either $f(x)$ or $g(x) \rightarrow \infty$, the other function $\rightarrow -\infty$. This is because the bigger a number you have as an outp{}ut for one function, the smaller the out{}put of the other function will be when you take its reciprocal. If one function has the outp{}ut 0, then the function that is its recipocal will be unde{}fined at that point because you can't divide anything by 0. Finally, if you have an intersection of a function and its reciprocal function, the points where they intersect will be at $x=1$ and $x=-1$ because the only values that are the same when divided into 1 are 1 and -1.

Using this information, we can sketch the graph of $f(x)=\frac{1}{g(x)}$. If this is the graph of $f(x)$: ${} math math {}$ We can see that ${} math math {}$ is the graph of $f(x)$ and $g(x)$. The graph of g(x) has asymptotes at the places were the graph of f(x) touches the x-axis because taking the reciprocal of those places would involve dividing by 0. Note that the places where the two graphs are at $x=1$ and $x=-1$ because $g(x)=1=\frac{1}{f(x)}$.

To find the domain of a reciprocal function like $g(x)$, it is all numbers for which $f(x)\neq 0$, because, again, that would involve dividing by 0. For example, if you had the function $f(x)=x^2+2x-8=(x-2)(x+4)$, which would show that the domain of g(x) would be all real numbers, $x\neq 2$ and $x\neq -4$.

We also covered how to find asymptotes and discontinuities on graphs. If $f(x)=\frac{x-3}{x^2-16}$ and $g(x)=\frac{x-4}{x^2-16}$, both are unde{}fined at $x=4$ and $x=-4$ because plugging in $x=4$ would put $0$ in the denominator. However, $g(x)$ will have a blank space at $x=4$ because it has 0 in the numerator as well as the denominator. This is shown in the math: $g(x)=\frac{x-4}{x^2-16}$ Plugging in $x=4$, we get $g(x)=\frac{4-4}{4^2-16}=\frac{0}{0}$ Therefore, the graph of $f(x)$ looks like this: ${} math math {}$ and the graph of g(x) looks like this, with a removable discontinuity at x=4: ${} math math {}$The formulas for the equations must then inc{}lude the note that they are unde{}fined at $x=4$, but they are also unde{}fined at $x=-4$ because $-4^2-16$ as a denominator also equals zero. Hence, we write them like this: $f(x)=\frac{x-3}{x^2-16} \forall x\neq 4, -4$ and $g(x)=\frac{x-4}{x^2-16} \forall x\neq 4, -4$. ${} math