Inverse functions, logarithms and trigonometric functions

This week has three separate topics, but all of them hang on one idea: undoing an operation. The logarithm undoes the exponential function, and $\arcsin$ undoes sine. In week 7 you will differentiate these functions, and in weeks 12-14 you will integrate with them. Someone who is not solid here can differentiate correctly but cannot simplify the result. The short list:

1. For an inverse function, write $y=f(x)$, isolate $x$, swap the letters.
2. Domain of the inverse = range of the original.
3. Logarithm rules turn a product into a sum and an exponent into a coefficient. They do not turn a sum into a product.
4. Check every root you get in a log equation. The argument of a logarithm must be positive.
5. Inverse trigonometric functions give values from a single interval (principal value).

Inverse function

If $f$ is one-to-one, meaning different $x$ go to different $y$, it has an inverse: $f^{-1}(y)=x \iff f(x)=y$. Test on the graph: if every horizontal line crosses the graph at most once, the function is one-to-one (horizontal line test).

Two properties of the inverse function:

$f\big(f^{-1}(x)\big)=x,\qquad f^{-1}\big(f(x)\big)=x$

Its graph is the reflection of the graph of $f$ in the line $y=x$. Domain and range swap places.

Example 1. $f(x)=3x-5$: $y=3x-5 \Rightarrow x=\dfrac{y+5}{3}$, so $f^{-1}(x)=\dfrac{x+5}{3}$.

Example 2. $f(x)=x^3+2$: $x=\sqrt[3]{y-2}$, so $f^{-1}(x)=\sqrt[3]{x-2}$. The cube is one-to-one everywhere.

$f(x)=x^2$ is not one-to-one on $\mathbb{R}$ ($f(2)=f(-2)$). If you restrict the domain to $x\ge 0$, its inverse is $\sqrt{x}$. If you see a restriction like “with domain $x\ge 2$” on an exam, it tells you which of the $\pm$ signs to choose.

Inverse values without a formula

Sometimes the formula for the inverse cannot be derived, but a single value is asked. $f(x)=x^3+x$ is one-to-one (increasing), but solving $y=x^3+x$ for $x$ is hard. The question “What is $f^{-1}(2)$?” is easy though: look for the $x$ that makes $f(x)=2$. If you try, $f(1)=1+1=2$, so $f^{-1}(2)=1$. In week 7, in “find $(f^{-1})'(a)$” questions, this is always the first step.

The most practical way to show that a function is one-to-one is to show that it is increasing or decreasing. A continuously increasing function cannot take the same value at two different points. You will see the increasing test with the derivative in week 9; for now, look at the graph or use the reasoning $x_1<x_2 \Rightarrow f(x_1)<f(x_2)$.

PITFALL · $f^{-1}(x)\ne \dfrac{1}{f(x)}$. The $-1$ on top is inverse function notation, not a power. For $\dfrac{1}{f(x)}$ write $[f(x)]^{-1}$.

Logarithm

For $a>0$, $a\ne 1$:

$\log_a x=y \iff a^y=x$

So $\log_a x$ is the answer to the question “to what power must I raise $a$ to get $x$?” The natural logarithm is $\ln x=\log_e x$. Its domain is $(0,\infty)$ and its range is $\mathbb{R}$. $\ln 1=0$, $\ln e=1$.

Example 3. $\log_3 54-\log_3 2=\log_3 27=3$.

Example 4. $\log_2(x+3)+\log_2(x-1)=5$.

log2[(x+3)(x-1)] = 5          product rule
(x+3)(x-1) = 2^5 = 32         apply the definition
x^2 + 2x - 35 = 0
(x+7)(x-5) = 0   ->  x = 5 or x = -7
x = -7: log2(-4) is undefined     DISCARD
answer: x = 5

Example 5. $3^x=7$. You cannot equalize the bases to 3. Take $\ln$ of both sides: $x\ln 3=\ln 7$, $x=\dfrac{\ln 7}{\ln 3}\approx 1.7712$.

Example 5b. $e^{2x}-5e^x+6=0$. This is a hidden quadratic equation: let $u=e^x$, $u^2-5u+6=0$, $(u-2)(u-3)=0$. $e^x=2 \Rightarrow x=\ln 2$, $e^x=3 \Rightarrow x=\ln 3$. Both roots are valid, because the $u$ values are positive.

General strategy table:

Since $\ln x$ and $e^x$ are inverses of each other, their graphs are symmetric about the line $y=x$. If $e^x$ has the horizontal asymptote $y=0$, then $\ln x$ has the vertical asymptote $x=0$. If $e^x$ passes through $(0,1)$, then $\ln x$ passes through $(1,0)$.

PITFALL · log rules that do not exist. $\ln(x+y)\ne \ln x+\ln y$ and $\dfrac{\ln x}{\ln y}\ne \ln\dfrac{x}{y}$. There is no rule for the logarithm of a sum.

Trigonometric functions

In calculus angles are always in radians: $180^\circ=\pi$. On the unit circle, the endpoint of the angle $\theta$ is $(\cos\theta,\sin\theta)$. From this:

$\tan\theta=\frac{\sin\theta}{\cos\theta},\quad \sec\theta=\frac{1}{\cos\theta},\quad \csc\theta=\frac{1}{\sin\theta},\quad \cot\theta=\frac{\cos\theta}{\sin\theta}$

Values you need to know

For other angles use the reference angle + quadrant sign. The functions that are positive in each quadrant: all in I, sin in II, tan in III, cos in IV.

Example 6. $\sin\tfrac{5\pi}{6}=\tfrac12$ (quadrant II, reference angle $\tfrac{\pi}{6}$). $\cos\tfrac{4\pi}{3}=-\tfrac12$ (quadrant III). $\tan(-\tfrac{\pi}{3})=-\sqrt3$ ($\tan$ is an odd function).

Basic identities

$\sin^2 x+\cos^2 x=1,\qquad 1+\tan^2 x=\sec^2 x$

$\sin 2x=2\sin x\cos x,\qquad \cos 2x=\cos^2 x-\sin^2 x=1-2\sin^2 x=2\cos^2 x-1$

$\sin(a\pm b)=\sin a\cos b\pm\cos a\sin b,\qquad \cos(a\pm b)=\cos a\cos b\mp\sin a\sin b$

The three forms of $\cos 2x$ are lifesavers in the integral weeks: $\sin^2 x=\tfrac{1-\cos 2x}{2}$, $\cos^2 x=\tfrac{1+\cos 2x}{2}$.

Example 7. On $[0,2\pi)$, $2\cos x+\sqrt3=0$. $\cos x=-\tfrac{\sqrt3}{2}$. The reference angle is $\tfrac{\pi}{6}$, and cosine is negative in quadrants II and III: $x=\pi-\tfrac{\pi}{6}=\tfrac{5\pi}{6}$ and $x=\pi+\tfrac{\pi}{6}=\tfrac{7\pi}{6}$.

Inverse trigonometric functions

Sine is not one-to-one, so we restrict the domain to an interval and take the inverse. These intervals are fixed and must be memorized:

Example 8. $\arcsin\big(\sin\tfrac{5\pi}{6}\big)$. The answer is not $\tfrac{5\pi}{6}$. $\sin\tfrac{5\pi}{6}=\tfrac12$, and $\arcsin\tfrac12=\tfrac{\pi}{6}$, because $\tfrac{5\pi}{6}$ is not in the range of arcsin.

Example 9. $\tan\big(\arcsin\tfrac13\big)$. $\theta=\arcsin\tfrac13$: opposite 1, hypotenuse 3, adjacent $\sqrt{9-1}=2\sqrt2$. $\tan\theta=\dfrac{1}{2\sqrt2}=\dfrac{\sqrt2}{4}$. This “draw a triangle” technique will come back in the inverse trigonometric substitution in week 14.

The same technique gives general formulas that help you simplify derivative results in week 7:

$\sin(\arccos x)=\sqrt{1-x^2},\qquad \cos(\arcsin x)=\sqrt{1-x^2},\qquad \sec(\arctan x)=\sqrt{1+x^2}$

See why the sign is always $+$ from the ranges: $\arccos x\in[0,\pi]$ and sine is never negative on this interval. $\arcsin x\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$ and cosine is never negative on this interval.

One more identity to know: for every $x\in[-1,1]$

$\arcsin x+\arccos x=\frac{\pi}{2}$

Check: for $x=\tfrac12$, $\tfrac{\pi}{6}+\tfrac{\pi}{3}=\tfrac{\pi}{2}$.

Common mistakes