Functions, transformations and exponential functions

The first week of Calculus is the one students skip with “I already know this” and then lose points on the midterm. Limit, derivative and integral questions are all asked about some function. A student who finds the domain wrongly, reverses a composition or shifts a graph the wrong way carries the same mistake into later weeks. These five rules will help you most on the exam:

1. The domain is found with three prohibitions: the radicand cannot be negative, the denominator cannot be zero, and the argument of a logarithm must be positive.
2. A composition $f(g(x))$ is read from the inside out. First $g$, then $f$.
3. A change inside the parentheses moves the graph horizontally, and a change outside moves it vertically. Inside, the sign works backwards: $f(x-3)$ goes right.
4. Even/odd test: compute $f(-x)$ and compare it with $f(x)$ or $-f(x)$.
5. In an exponential equation, equalize the bases. If that fails, let $u=a^x$.

Function, domain, range

A function is a rule that assigns exactly one value $y$ to every $x$ in the domain. The domain is the set of $x$ for which the formula makes sense. The range is the set of $y$ values the function actually takes.

If the question does not give a domain, it is asking for the “natural domain”. To find it, scan the formula and remove the forbidden points:

Example 1. The domain of $f(x)=\dfrac{\sqrt{x+3}}{x^2-4}$.

Radicand: $x+3\ge 0 \Rightarrow x\ge -3$. Denominator: $x^2-4\ne 0 \Rightarrow x\ne \pm 2$. The intersection of the two:

$[-3,-2)\cup(-2,2)\cup(2,\infty)$

$x=-3$ is included because $\sqrt{0}$ is defined. When writing intervals, a square bracket means “included” and a round parenthesis means “excluded”.

PITFALL · simplifying does not change the domain. $\dfrac{x^2-1}{x-1}$ simplifies to $x+1$, but $x=1$ is still forbidden. Always find the domain from the original formula.

Finding the range

There is no formula for the range; you look at the structure of the function. The two most common techniques:

• Complete the square. $f(x)=x^2-4x+7=(x-2)^2+3$. A square is at least 0, so $f(x)\ge 3$. The range is $[3,\infty)$, and the minimum occurs at $x=2$.
• Build from a known piece. $e^{-x}$ takes every positive number. So $3-2e^{-x}$ takes $(-\infty,3)$. Transform the interval at each step: multiplying multiplies the endpoints (multiplying by a negative flips the interval), and adding shifts the endpoints.

The most reliable way to test whether a value $y_0$ is in the range is to try to solve $f(x)=y_0$ for $x$. If there is a solution, $y_0$ is attained. You will use the same idea in week 2 when finding inverse functions.

Piecewise functions

Functions that use a different formula on different intervals. When evaluating, first check which interval $x$ lies in, then use the formula of that piece. The absolute value is also piecewise:

$|x|=\begin{cases} x, & x\ge 0\\ -x, & x<0\end{cases}$

Composite functions (composition)

$(f\circ g)(x)=f(g(x))$. First $g$ is applied, and the result goes into $f$. In general $f\circ g\ne g\circ f$.

Example 2. $f(x)=x^2-1$, $g(x)=\sqrt{x}$.

$(f\circ g)(x)=f(\sqrt{x})=(\sqrt{x})^2-1=x-1,\qquad x\ge 0$

$(g\circ f)(x)=g(x^2-1)=\sqrt{x^2-1},\qquad |x|\ge 1$

For $f\circ g$ the formula came out as $x-1$, but the domain is not $\mathbb{R}$. If you plug in $x=-4$, you first have to compute $\sqrt{-4}$, which is undefined. The domain of a composition: the $x$ where $g$ is defined and $g(x)$ falls in the domain of $f$.

It is also asked in the opposite direction: “Write $F(x)=\sqrt{x^2+5}$ as $f\circ g$.” The answer is $g(x)=x^2+5$, $f(x)=\sqrt{x}$. This decomposition is the basis of the chain rule in week 6.

Even and odd functions

Example 3. Test three functions.

• $f(x)=\dfrac{x}{x^2+1}$: $f(-x)=\dfrac{-x}{x^2+1}=-f(x)$. Odd.
• $g(x)=x^4-3x^2$: $g(-x)=x^4-3x^2=g(x)$. Even.
• $h(x)=x^2+x$: $h(-1)=0$, $h(1)=2$. Neither $h(-1)=h(1)$ nor $h(-1)=-h(1)$. Neither odd nor even.

A single counterexample is enough to say “neither odd nor even”. To say “odd” or “even”, you must show the equality for a general $x$.

Shortcut: odd × odd = even, even × even = even, odd × even = odd. The same holds for division.

Graph transformations

If you know the graph of $y=f(x)$, you can sketch these from memory ($c>0$):

The logic: whatever is inside the parentheses acts on $x$. In the graph of $f(x-3)$, the value that used to be at $x=0$ is now seen at $x=3$. That is why the minus sign inside takes the graph to the right.

Example 4. Obtain the graph of $y=3-2^{x+1}$ from $y=2^x$.

y = 2^x          base graph, asymptote y = 0, passes through (0,1)
y = 2^(x+1)      1 unit to the left
y = -2^(x+1)     reflection in the x-axis, the graph drops below the axis
y = 3 - 2^(x+1)  3 units up, asymptote y = 3

$y$-intercept: $3-2^1=1$. The range is $(-\infty,3)$. The asymptote also moved up 3 units: $y=3$.

PITFALL · order matters. “First stretch by 2, then shift up 1” and “first shift up 1, then stretch by 2” give different graphs: $2f(x)+1$ versus $2(f(x)+1)=2f(x)+2$. Apply the steps in the order given in the question.

Exponential functions

$f(x)=a^x$, where $a>0$ and $a\ne 1$. Properties:

• The domain is $\mathbb{R}$ and the range is $(0,\infty)$. It is never zero or negative.
• It passes through $(0,1)$, because $a^0=1$.
• It is increasing if $a>1$ and decreasing if $0<a<1$. $y=0$ is a horizontal asymptote.

Exponent rules:

$a^{x+y}=a^x a^y,\qquad a^{x-y}=\frac{a^x}{a^y},\qquad (a^x)^y=a^{xy},\qquad (ab)^x=a^x b^x,\qquad a^{-x}=\frac{1}{a^x}$

The natural base e

The number $e$ is the following limit:

$e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\approx 2.71828$

For $n=100$, $(1.01)^{100}\approx 2.7048$; as $n$ grows, it settles at $2.71828$. In calculus almost every exponential function is written with $e^x$, because, as you will see in week 7, the derivative of $e^x$ is itself.

Solving exponential equations

Method 1: equalize the bases. $a^{A}=a^{B} \Rightarrow A=B$.

Example 5. $9^x=27^{x-2}$.

$3^{2x}=3^{3(x-2)} \Rightarrow 2x=3x-6 \Rightarrow x=6$

Method 2: substitution. If the equation contains both $a^{2x}$ and $a^x$, let $u=a^x$ and a quadratic equation appears. Since $u>0$, discard negative roots.

Example 6. $2^{2x}-6\cdot 2^x+8=0$. $u=2^x$: $u^2-6u+8=0$, $(u-2)(u-4)=0$. $2^x=2 \Rightarrow x=1$, $2^x=4 \Rightarrow x=2$.

If the bases cannot be equalized (like $5^x=2^{x+1}$), you need logarithms. That is the topic of week 2.

Growth model

The sentence “it doubles every $T$ hours, starting at $N_0$” translates to the model:

$N(t)=N_0\cdot 2^{t/T}$

For half-life, write $\tfrac12$ instead of $2$.

Common mistakes