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# 5.6: Transcendental Functions

Difficulty Level: 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 f(x)=x2\begin{align*}f(x)=x^2\end{align*}? The lesson in the text illustrates clearly that it is not a one-to-one function. The instructions for finding an inverse function state to solve for the dependent variable, which states that the inverse function is ±x\begin{align*}\pm \sqrt{x}\end{align*}, which is really not a function at all, with two outputs for every input. If we take only the positive part, then we can call it a function. Now it's important to understand that this isn't a complete inverse, but is more of a functional inverse. There are many instances which the negative values are not needed, like in many physical problems involving time, distance or other quantities that can't logically have negative values.

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 e\begin{align*}e\end{align*}. The common definition is the one listed in the text: limn(1+1n)n\begin{align*}\lim_{n \to \infty} \left (1 + \frac{1}{n} \right )^n\end{align*}. There are a couple of different ways to find this quantity, some of which make good problems for students.

If a person makes a \$1\begin{align*}\1\end{align*} investment in a bank that pays 100%\begin{align*}100\%\end{align*} interest per year, how much is in the bank at the end of the year? If the interest in compounded at two points in the year, how much is in the bank? How about if the interest is compounded quarterly? Monthly? Daily? Every second? What is the maximum amount that can be in the bank at the end of the year.

The only thing to remember here is the compound interest formula: A=p(1+rt)yt\begin{align*}A = p \left (1 + \frac{r}{t} \right )^{yt}\end{align*} where p\begin{align*}p\end{align*} is the principle amount, r\begin{align*}r\end{align*} is the periodic rate, y\begin{align*}y\end{align*} is the number of periods, and t\begin{align*}t\end{align*} is the number of times per period the interest is compounded. By plugging in the information for each question, it should become clear that the amount is a sequence approaching 2.71828\begin{align*}2.71828\end{align*}, with the final question resembling the limit expressed above.

Another, seemingly unrelated, way to find the number is with a classic gambling question. If there is a slot machine that hits every 1\begin{align*}1\end{align*} out of n\begin{align*}n\end{align*} times, and a person plays the machine n\begin{align*}n\end{align*} times. What is the probability the player does not win anything if n=10?n=100?n=1,000,000?n\begin{align*}n=10? n=100? n=1,000,000? n\end{align*} goes to \begin{align*}\infty\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: (nk)pk(1p)nk\begin{align*}\binom{n}{k} p^k (1-p)^{n-k}\end{align*} where n\begin{align*}n\end{align*} is the number of trials, k\begin{align*}k\end{align*} is the number of successes, and p\begin{align*}p\end{align*} is the probability of success. Plugging in the first question looks like: (100)(110)0(1110)10=1×1×9101010.3487\begin{align*}\binom{10}{0} \left (\frac{1}{10} \right )^0 \left (1- \frac{1}{10} \right )^{10} = 1 \times 1 \times \frac{9^{10}}{10^{10}} \approx .3487\end{align*}. Skipping the rest, the last is: limn(n0)(11n)n=limn(11n)n\begin{align*}\lim_{n \to \infty}\binom{n}{0} \left (1 - \frac{1}{n} \right )^n = \lim_{n \to \infty}\left (1 - \frac{1}{n} \right )^n\end{align*} which looks an awful lot like the limit for e\begin{align*}e\end{align*}. In fact, this is equal to e1\begin{align*}e^{-1}\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 dydx=p(x)q(y),q(y)=0\begin{align*}\frac{dy}{dx} = p(x)q(y),q(y)\cancel {=} 0\end{align*}. Letting a new function r(y)=1q(y)\begin{align*}r(y) = \frac{1}{q(y)}\end{align*} then the differential equation can be rewritten as dydx=p(x)p(y)\begin{align*}\frac{dy}{dx} = \frac{p(x)}{p(y)}\end{align*}. Multiplying both sides by r(y)\begin{align*}r(y)\end{align*} yields r(y)dydx=p(x)\begin{align*}r(y) \frac{dy}{dx} = p(x) \end{align*}. Now integrate both sides with respect to x\begin{align*}x\end{align*}: r(y)dydxdx=p(x)dx\begin{align*}\int r(y)\frac{dy}{dx}dx = \int p(x)dx\end{align*}. Which means r(y)dy=p(x)dx\begin{align*}\int r(y)dy = \int p(x) dx\end{align*} will give the solution when the integrals are taken. Notice that it sure looks like we are cross multiplying when the intermediate steps are not considered, but it is not exactly the case. Some students may think of it that way, which isn't a bad too to remember what to do to find a solution, but there are some things that aren't helped by thinking of it that way. Note that the original isn't a fraction on both sides, so that it is not a necessity for separable equations. For instance: dydx=xy2cos(y2)\begin{align*}\frac{dy}{dx} = x y ^ 2 \cos (y ^ 2)\end{align*} is a separable equation. Sometimes it is helpful to rewrite the right hand side as a fraction to keep the process consistent.

## 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 1udu=lnu\begin{align*}\int \frac{1}{u}du = \mathrm {ln} u\end{align*}. The process of integration is beginning to get drilled in, students know that they should convert denominators to negative exponents in the numerator if possible and then apply the reverse-power rule to find the anti-derivative. The problem is this process will not work for special trig integrals and log integrals. Furthermore, there are few clues that can help the student along. For example:

x1+x2\begin{align*}\int \frac{x}{1 + x ^ 2}\end{align*} is solvable by substitution but x2+2x1+x2dx\begin{align*}\int \frac{x ^ 2 + 2x}{1 + x ^ 2}dx\end{align*} can't, and the anti-derivative that gives an answer of arctan will need to be used. There will be other problems where using the method of partial fractions works for a rational expression, and other where completing the square and using a trig substitution works, and the problems look nearly identical. Here is my recommendation for attempting to solve these problems.

• 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: x1x2dx\begin{align*}\int \frac{x}{\sqrt{1 - x ^ 2}}dx\end{align*} 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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