Last week we observed that \(e^x\) is the inverse of the natural log function \(\ln x\). In fact, for any base \(b\), \(\log_b(x)\) and \(b^x\) are inverses. The inverse of a function \(f(x)\) is denoted \(f^{-1}(x)\) which is terrible notation, because we also use the notation \(x^{-1}\) to denote the reciprocal. The two meanings are not the same, \[f^{-1}(x) \ne \frac{1}{f(x)} ~~~\text{even though}~~~ (f(x))^{-1} = \frac{1}{f(x)}.\] On behalf of all mathematicians, let me apologize to you for this terrible notation!
Inverse functions cancel, that is: \[f^{-1}(f(x)) = x ~~~ \text{and} ~~~ f(f^{-1}(x)) = x.\]
A function is invertible if each \(y\)-value only occurs once on the graph, that is, if the function passes the horizontal line test. Such functions are called one-to-one functions (which really means that for every two different x-values, you get two different y-values, so they should be called two-to-two functions!).
We pointed out that a continuous function that is always increasing is automatically invertible (and so are continuous decreasing functions). You can tell if a function is always increasing or decreasing by looking at the derivative.
Lots of important functions like \(y=x^2\) and \(y = \sin x\) are not invertible, because they fail the horizontal line test. To deal with these functions, we restrict the domain to a subset of the x-values where the function is invertible. For example, \(y= x^2\) is invertible on the domain \([0,\infty)\). Then the inverse is \(y = \sqrt{x}\). Likewise, \(\sin x\) is invertible as long as \(x\) is between \([-\pi/2,\pi/2]\).
We reviewed how to verify that two functions are inverses by plugging one function into the other with this example:
We also reviewed how to find the inverse of a function \(y = f(x)\) by solving for \(x\) as a function of \(y\) with these two examples:
Find the inverse of \(f(x) = x^5 + 2\).
Invert the function \(y = \frac{4x-1}{2x+3}\). (This one was tricky, because you need to use algebra to combine the like terms in order to solve for \(x\)).
We also did some conceptual problems:
We finished today by talking about inverse trig functions. The most important thing to know is that regular trig functions input an angle and output a ratio. That’s why we write \(\sin \theta\), since the letter \(\theta\) usually represents an angle. Inverse trig functions are the reverse, they input a ratio, and output an angle.
Notation: We’ll write \(\arcsin x\) and \(\sin^{-1}(x)\) both to represent the inverse sine (and likewise, there is an arc cosine, tangent, etc.
Today we talked about how to compute inverse trig functions using reference triangles. We did the following examples:
\(\sec^{-1}( \sqrt{2} )\)
\(\arctan (\sqrt{3} )\)
Then we demonstrated how to use implicit differentiation and reference triangles to differentiate inverse trig functions. We showed that \(\displaystyle \frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}\) by first writing: \[y = \sin^{-1}(x)\] Then we solved for \(x\) by taking the sine of both sides: \[\sin(y) = x\] Then we differentiated both sides (with respect to x): \[\cos(y) y' = 1\] So \[y' = \sec(y).\] Then to simplifty \(\sec(y) = \sec(sin^{-1}(x))\) we use a reference triangle to get the final result.
We usually won’t use a reference triangle to find the derivatives of inverse trig functions. Instead, all of the derivatives of the inverse trig functions are on the formula sheet. Use the formulas on the formula sheet to find:
\(\displaystyle \frac{d}{dx} \cos^{-1}(4x^2)\).
\(\displaystyle \frac{d}{dx} \sin^{-1}\left(\frac{x}{a}\right)\) where \(a\) is a positive constant.
The solution to #5 is \(\displaystyle \frac{1}{a} \frac{1}{\sqrt{1-(x/a)^2}}\). You don’t have to simplify, but you can simplify it to \(\displaystyle \frac{1}{\sqrt{a^2 - x^2}}\) which looks a lot nicer. Notice, that this is one of the functions listed on the formula sheet for integrals.
As always, you should be able to use derivatives to find max/min values. For example,
Today we introduced differential equations. A differential equation is an equation with a derivative or differentials in it. We did several examples:
Exponential growth (and decay) \(\displaystyle \frac{dP}{dt} = k P.\) In English, this differential equation literally means the rate of growth of a population \(P\) is directly proportional to the size of the population. The constant \(k\) is called the proportionality constant.
Explosion equation \(\displaystyle \frac{dy}{dt} = k y^2\). This equation models a chemical reaction where the rate of growth of a product \(y\) of a reaction is directly proportional to the amount of the product squared. When we solved this equation, we saw that it leads to infinite growth in a finite amount of time!
To solve a differential equation, you need to separate the variables (see section 6.5 in the textbook) and then integrate. We also talked about how you can graph a differential equation by plotting the slope field (see Examples for how to draw these using Sage). This lead to 3 important ideas:
Idea 1. A solution of a differential equation is a function, not a number. Every differential equation has infinitely many different solutions.
Idea 2. The solution functions all follow the slope field.
Idea 3. To find a particular solution, use the initial conditions to solve for the constant \(C\).
In class we solved a couple differential equations including:
\(\displaystyle \frac{dy}{dx} = \frac{\sin x}{y}\).
\(\displaystyle \frac{dP}{dt} = 2 P\).
\(\displaystyle \frac{dy}{dt} = ky^2\).
We also looked at the slope field for a logistic equation \(\displaystyle \frac{dP}{dt} = 1.05 P \left(1-\frac{P}{500} \right).\) Even though this equation is really hard to solve, we can still predict what the solution functions will look like since they have to follow the slope field.