# 5.6: Transcendental Functions

**At Grade**Created by: CK-12

## Inverse Functions

A problem that is worth thinking about as a useful tool, as well as a foreshadowing of future ideas, is how to work around the one-to-one restriction. There are times when an inverse is needed, even though the function is not one-to-one, and there may be some restrictions that can be applied to make it happen. For instance: what is the inverse function \begin{align*}f(x)=x^2\end{align*}

Another key consideration is what domain restrictions need to be made. Here, the range becomes the domain, which needs to be explicitly stated as many functions will not have a range of all reals. For the case of our example, the domain of the inverse is all non-negative real numbers.

Inverse functions can introduce many technical problems. They should always be treated with careful attention, as the problems are often not immediately apparent.

## Differentiation and Integration of Logarithmic and Exponential Functions

Here it can be entertaining to take a number of different looks at \begin{align*}e\end{align*}

If a person makes a \begin{align*}\$1\end{align*}

The only thing to remember here is the compound interest formula: \begin{align*}A = p \left (1 + \frac{r}{t} \right )^{yt}\end{align*}

Another, seemingly unrelated, way to find the number is with a classic gambling question. If there is a slot machine that hits every \begin{align*}1\end{align*}

If students have not had a course in probability and statistics, they may not be familiar with how to find this probability. Since the outcomes are either win or lose, this is a binomial probability: \begin{align*}\binom{n}{k} p^k (1-p)^{n-k}\end{align*}

## Exponential Growth and Decay

The first time students see the separation of variables it can cause some confusion. Leibniz introduced the differential notation that we use specifically for the purposes of treating the individual parts of the differentials like they are variables. To understand why separable problems work, it may be useful to look at the justification for the general solution method.

Assume a differential equation can be written as \begin{align*}\frac{dy}{dx} = p(x)q(y),q(y)\cancel {=} 0\end{align*}

## Derivatives and Integrals Involving Trigonometric Functions

Using the stranger of the trig integrals is one of the toughest integration techniques. It is not unlike the challenge faced when trying to remember the integral of \begin{align*}\int \frac{1}{u}du = \mathrm {ln} u\end{align*}

\begin{align*}\int \frac{x}{1 + x ^ 2}\end{align*}

- Go ahead and try the u-substitution or the easier method. Still a large majority of the problems students are going to encounter use the basic integration techniques. One small problem is that you want students to feel confident with their work so they can tell the difference between reaching a dead end and just being stuck, or having made a mistake. Still, no reason to try to out-think the problem and go straight to the trig integral.

- If there is no other variable outside the denominator, or outside the radical in the denominator, then it is likely to be a trig antiderivative. For example: \begin{align*}\int \frac{x}{\sqrt{1 - x ^ 2}}dx\end{align*}
∫x1−x2−−−−−√dx does not have a trig antiderivative, but \begin{align*}\int \frac{x}{\sqrt{1 - x ^ 2}}dx\end{align*} does. That extra variable makes the chain rule part of substitution possible.

- Have visual reminders up for students for a long period of time. Students need to see the form of the trig antiderivatives frequently to help commit them to memory. Problems may not always be presented in exactly the form that has a known antiderivative, so recognizing problems that are close to the form, and then using algebra to make it work, is dependent on having those forms committed to memory.

## L'Hopitals Rule

Why does l'Hopital's rule work? It possibly seems odd to be able to take a derivative of a part of a function to help find a limit. Students are likely apt to accept the rule without reason as it makes a number of challenging limits much easier to compute. l'Hopitals rule is a result of local linearity of functions.

\begin{align*}\lim_{x \to n}\frac{f(x)}{g(x)} = \lim_{x \to n}\frac{\frac{f(x)}{x - n}} {\frac{g(x)}{x - n}} = \frac{f'(x)}{g'(x)} = \lim_{x \to n}\frac{f'(x)}{g'(x)}\end{align*}

Some places will present that reasoning as proof, but it is not really proof. The actual proof of the rule comes from examining each of the indeterminate forms individually and then applying the mean value theorem. This short piece of reasoning is, however, a good examination of what is going on with the local linearity. Put in English, if we were to examine the lines tangent to each individual curve at the place where the limit is to be taken, then the ratio of slopes is going to be a good approximation of the ratio of the original functions. This is most clearly illustrated in the zero over zero indeterminate form.

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