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1. A Trip Down Memory Lane...

Table of Contents | 2. Exponents and Logarithms ≫

Since it’s the first week of the quarter, it’s a great opportunity to recap some helpful formulae and techniques from the first quarter of calculus (Math 31A) and precalculus, which you may have taken a while ago. These skills will be consistently used throughout this quarter, so it’s important that you’re up to speed and that you have a refreshed memory!

The Domain, Range, and Inverse

The domain of a function \(f(x)\) is the set of all “reasonable” values of \(x\) that can be input into the function. When I say “reasonable”, I mean that this value of \(x\) will not cause you to divide by \(0\) or take the squareroot of a negative number, or perform any other affront to decency. For instance:

• The domain of \(f(x) = \sin x\) is all real numbers. I can plug anything I want into \(\sin\) and get a number out the other end.
• The domain of \(g(x) = \frac{1}{x}\) is all nonzero real numbers. \(0\) is not in the domain, since \(x=0\) would cause me to divide by \(0\).
• The domain of \(h(x) = \sqrt{5-x}\) is \(x \leq 5\). No numbers \(x>5\) are in the domain because that would make me take a squareroot of a negative number.

Sometimes, we will artificially restrict the domain of a function, such as in “\(f(x) = 2x+3\) with the domain \(\left[ 1,2 \right]\)”. This means I am only allowing \(x\) to take values between \(1\) and \(2\) (inclusive), and I am forbidden from using other values of \(x\) (for no good reason at all).

The range of a function \(f(x)\) is the set of all possible outputs of \(f(x)\). For instance,

• The range of \(f(x) = \sin x\) is \(\left[ -1,1 \right]\). It takes every value between \(-1\) and \(1\) (inclusive), but it doesn’t take any values outside of that.
• The range of \(g(x) = \frac{1}{x}\) is all nonzero real numbers. There is no value of \(x\) for which \(g(x) = 0\).
• The range of \(h(x) = \sqrt{5-x}\) is \(\left[ 0,\infty \right]\). The squareroot function only selects the positive squareroot, and each of these is the squareroot of some number.

Finally, a function \(f(x)\) has an inverse if it is one-to-one: no two inputs \(x\) and \(y\) in its domain can have the same output. That is, if \(x\neq y\) are both in the domain of \(f\), then \(f(x)\neq f(y)\). Graphically, you can check this using the horizontal line test. If you can draw a horizontal line that intersects the graph of \(f\) more than once, then \(f(x)\) does not have an inverse.

If you know \(f(x)\) has an inverse and you know the formula for \(f\), you can find a formula for the inverse algebraically as follows. First, write down the equation \(y = f(x)\), then expand the formula for \(f\). Finally, isolate \(x\) as a function of \(y\). The domain of the inverse is restricted to the range of \(f\), whatever it may be.

• \(f(x) = \sin x\) does not have an inverse, since it crosses the \(x\)-axis (a horizontal line) many, many times.
• \(g(x) = \frac{1}{x}\) does have an inverse: every horizontal line crosses the graph of \(g(x)\) at most once. To find its inverse, we set up \(y = f(x) = \frac{1}{x}\). Isolating \(x\) gives \(x = \frac{1}{y}\), so the inverse is \(g ^{-1}(y) = \frac{1}{y}\).
• \(h(x) = \sqrt{5-x}\) also has an inverse, since horizontal lines cross the graph of \(h\) at most once. To find the inverse, we set up \(y = \sqrt{5-x}\). Squaring both sides, we have \(y^2 = 5-x\). Isolating \(x\) gives \(x = 5 - y^2\), so \(h ^{-1}(y) = 5-y^2\), restricted to the domain \([0,\infty)\).

Derivative and Integral Refresher

I probably won’t review the conceptual basis of derivatives and integrals, that takes way too long (and has entire courses dedicated to it). Instead, let’s rattle off some of the important formulae you will need to remember, starting with basic derivatives and integrals: \[\begin{aligned} \frac{d}{dx} x^n &= n\cdot x ^{n-1} \qquad& \int x^n dx &= \frac{x ^{n+1}}{n+1} + C, \\ \frac{d}{dx}\cos x &= -\sin x & \int \cos x\ dx &= \sin x + C,\\ \frac{d}{dx}\sin x &= \cos x & \int \sin x\ dx &= -\cos x + C, \\ \frac{d}{dx}e^x &= e^x & \int e^x dx &= e^x + C. \\ \end{aligned}\]

Integrals and derivatives distribute over addition and multiplication by real numbers. For instance, \[\frac{d}{dx} \left[ 5\cos x + 3x^2 \right] = \frac{d}{dx}(5\cos x) + \frac{d}{dx}\left( 3x^2 \right) = -5\sin x + 6x.\] Similarly, \[\int \left(7\sin x-3x^4\right) dx = 7\int \sin x\ dx - 3\int x^4\ dx = -7\cos x - \frac{3}{5}x^5 + C.\]

For more complicated combinations of functions, you will need to apply the product rule, chain rule, or quotient rule as necessary.

When differentiating a product of functions, differentiate each term in the product while leaving the rest of the terms alone, then add everything together: \[\frac{d}{dx}\left( x^2\sin x \right) = 2x\sin x + x^2\cos x.\] The \(2x\sin x\) comes from the derivative “hitting” the \(x^2\), and the \(x^2\sin x\) comes from the derivative “hitting” the \(\cos x\).

When differentiating nested functions, use the chain rule: differentiate the “outside” function without touching the “inside” function in the nest, then multiply in the derivative of the “inside” function (the “chain”). For instance, \[\frac{d}{dx} \sin\left( 5x^2+1 \right) = \cos \left( 5x^2+1 \right) \cdot 10x.\] The \(10x\) is the derivative of the inside guy, the \(5x^2+1\). Sometimes, the “outside” and “inside” can be hard to tell, but the “inside” function in the chain rule is generally whatever is in parentheses: \[\frac{d}{dx} \left( \cos x + x \right)^7 = 7\left( \cos x + x \right)^6 \cdot \left( -\sin x + 1 \right).\] Here, the “outside” is \(\left( \cdots \right)^7\), and the “inside” is \(\cos x + x\).

I personally don’t use the quotient rule very much — it has a bulky formula that’s annoying to keep track of. If you want it, here it is: \[\frac{d}{dx} \frac{f(x)}{g(x)} = \frac{f’(x) g(x) - f(x) g’(x)}{g(x)^2}.\]