The Lumenary · Grades 9–12

Functions and their graphs (precalculus)

Name: ______________________

Date: ______________________

You will learn to

Sketch the standard function families and their transformations $y = a\,f(b(x - h)) + k$ , and prove the transformation rule
Compose real functions and find the natural domain of a composite
Decide when a function has an inverse, find it on an interval where the function is injective, and prove that its graph is the reflection in $y = x$
Identify zeros, symmetry (even and odd functions), asymptotes and end behaviour of polynomial and rational functions

1.
How is the graph of $y = f(x + 3)$ obtained from the graph of $y = f(x)$ ?
- A.Move it 3 units to the left.
- B.Move it 3 units to the right.
- C.Move it 3 units up.
- D.Move it 3 units down.
2.
Let $f(x) = 2x + 1$ and $g(x) = x^2$ . Compute $(g \circ f)(2)$ .
3.
What is the natural domain of $h(x) = \dfrac{\sqrt{x - 1}}{x - 3}$ ?
- A. $x \ge 1$ and $x \ne 3$
- B. $x \ge 1$
- C. $x \ne 3$
- D. $x > 3$
4.
Let $f(x) = \dfrac{x - 1}{x + 2}$ for $x \ne -2$ . Which formula gives $f^{-1}(x)$ ?
- A. $\dfrac{2x + 1}{1 - x}$
- B. $\dfrac{x + 2}{x - 1}$
- C. $\dfrac{x + 1}{x - 2}$
- D.There is no inverse, because $f$ is not injective.
5.
Find all vertical asymptotes $x = a$ of $r(x) = \dfrac{x + 1}{x^2 - 1}$ . Enter the set of numbers $a$ , for example {2, 5}.
6.
Which of these functions are odd, meaning $f(-x) = -f(x)$ for every $x$ in the domain? Choose all that apply.
- A. $x^3 - x$
- B. $x^2 + 1$
- C. $\dfrac{1}{x}$
- D. $x^3 + 1$
- E. $|x|$
7.
The function $f(x) = \dfrac{1}{x}$ is strictly decreasing on $(-\infty, 0)$ and on $(0, \infty)$ . Which statement is true on its whole domain $x \ne 0$ ?
- A. $f$ is strictly decreasing on its whole domain.
- B. $f$ is injective on its whole domain, and it is its own inverse.
- C. $f$ is not injective, because it is not monotone on its whole domain.
- D. $f$ has no inverse, because $0$ is not in its range.
8.
Which Lean statement says that the graph of $y = f(x - h)$ is the graph of $f$ moved $h$ units to the right? Here graph f is the set of points (x, f x), and F '' S is the image of the set S under the map F.
- A.graph (fun x => f (x - h)) = (fun p => (p.1 + h, p.2)) '' graph f
- B.graph (fun x => f (x - h)) = (fun p => (p.1 - h, p.2)) '' graph f
- C.graph (fun x => f x - h) = (fun p => (p.1 + h, p.2)) '' graph f
- D.∀ x, f (x - h) = f x + h
9.
Write inverse_value(f, y, lo, hi, steps) that evaluates the inverse of a strictly increasing function f at y by bisection. Assume f(lo) <= y <= f(hi). Repeat steps times: look at the midpoint mid; if f(mid) <= y the answer is in [mid, hi], otherwise in [lo, mid]. Return the midpoint of the final interval.
Computational — Practice

Answer key — Functions and their graphs (precalculus)

1.
$f(x + 3) = f(x - (-3))$ , so by the transformation rule the graph moves by $h = -3$ , that is, $3$ units to the left. Check: the new function takes the value $f(0)$ at $x = -3$ .
2.
$f(2) = 5$ and $g(5) = 25$ , so $(g \circ f)(2) = 25$ . In the other order, $(f \circ g)(2) = f(4) = 9$ : composition is not commutative.
3.
We need $x - 1 \ge 0$ (real square root) and $x - 3 \ne 0$ (no division by zero). So the natural domain is $[1, 3) \cup (3, \infty)$ .
4.
Solve $y = \tfrac{x - 1}{x + 2}$ : $xy + 2y = x - 1$ , so $x(y - 1) = -1 - 2y$ and $x = \tfrac{2y + 1}{1 - y}$ (for $y \ne 1$ ). So $f^{-1}(x) = \tfrac{2x + 1}{1 - x}$ . The value $1$ , missing from the range of $f$ , is the horizontal asymptote of $f$ and the vertical asymptote of $f^{-1}$ .
5.
$r(x) = \tfrac{x + 1}{(x - 1)(x + 1)} = \tfrac{1}{x - 1}$ for $x \ne \pm 1$ . Near $x = 1$ , $|r(x)|$ grows without bound: a vertical asymptote. Near $x = -1$ , $r(x) \to -\tfrac12$ : the graph has a hole at $\left(-1, -\tfrac12\right)$ . So the only vertical asymptote is $x = 1$ .
6.
$x^3 - x$ and $\tfrac1x$ are odd. $x^2 + 1$ and $|x|$ are even. $x^3 + 1$ is neither: it splits as the even part $1$ plus the odd part $x^3$ . (An odd function defined at $0$ must have $f(0) = 0$ , which rules out $x^3 + 1$ at once.)
7.
$f$ is not decreasing on the whole domain ( $f(-1) < f(1)$ ), but it is injective, and $f(f(x)) = x$ , so $f$ is its own inverse. The lesson's theorem says strictly monotone $\Rightarrow$ injective. This example shows the converse fails when the domain is not an interval. (For a continuous function on an interval, injective does imply strictly monotone; see the topology viewpoint.)
8.
The correct statement is graph (fun x => f (x - h)) = (fun p => (p.1 + h, p.2)) '' graph f: the new graph is the image of the old graph under $(x, y) \mapsto (x + h, y)$ . It is proved in the lesson as graph_shift.
9.
```
def inverse_value(f, y, lo, hi, steps):
    for _ in range(steps):
        mid = (lo + hi) / 2
        if f(mid) <= y:
            lo = mid
        else:
            hi = mid
    return (lo + hi) / 2
```
The answer always stays in $[lo, hi]$ (because $f$ is increasing and continuous), and each step halves the width. After $n$ steps the error is at most $(hi - lo)/2^{n+1}$ , in exact arithmetic.