Rates of Change & Combining Functions: Test Review

Everything for the Unit 6 test on one page: the key facts from all 7 lessons, the mistakes that cost the most marks, and a practice test that points you back to the lessons you need.

Key facts

Average Rate of Change

Average rate of change on $[a, b]$ is $\frac{f ( b ) - f ( a )}{b - a}$ — the slope of the secant through $(a, f (a))$ and $(b, f (b))$ .
It equals $\frac{Δ y}{Δ x}$ : change in output divided by change in input.
Units of the rate are (output units) per (input unit), e.g. m/s, dollars per item, people per year.
A positive rate means the output rose overall; negative means it fell; zero means it ended where it began.
The order of subtraction must match in numerator and denominator — both "later minus earlier" or both "earlier minus later".

Instantaneous Rate of Change

Instantaneous rate of change at $x = a$ is the slope of the tangent line to $y = f (x)$ at $(a, f (a))$ .
The difference quotient $\frac{f ( a + h ) - f ( a )}{h}$ is the slope of the secant from $a$ to $a + h$ .
Shrinking $h$ makes the secant slope approach the tangent slope; $h$ may become tiny but never exactly $0$ .
Centred difference estimate: $\frac{f ( a + h ) - f ( a - h )}{2 h}$ — usually more accurate than a one-sided estimate with the same $h$ .
Algebraically: expand, cancel the common factor of $h$ , then discard the remaining terms containing $h$ because they vanish as $h$ becomes small.
Units of an instantaneous rate are the same as for an average rate: output units per input unit.

Rates of Change and the Shape of a Graph

Positive instantaneous rate on an interval: the function is increasing there. Negative: decreasing.
At a local maximum the rate changes from positive to negative; at a local minimum, from negative to positive. The tangent is horizontal at both.
A zero rate without a sign change, as for $y = x^{3}$ at $x = 0$ , is not a turning point.
The graph is steepest where the rate is largest in size, often at a point of inflection between turning points.
For a sinusoid the rate is zero at maximums and minimums and greatest in size at the axis, where it equals $∣ a ∣∣ k ∣$ .
For an exponential function the rate is proportional to the value: rate ÷ value is the same at every point.

Combining Functions

$(f \pm g) (x) = f (x) \pm g (x)$ and $(f g) (x) = f (x) \cdot g (x)$ ; the domain of each is $D_{f} \cap D_{g}$ .
$(\frac{f}{g}) (x) = \frac{f ( x )}{g ( x )}$ , with domain $D_{f} \cap D_{g}$ minus every $x$ where $g (x) = 0$ .
Domain restrictions are read off the original functions, never off the simplified result.
A function is even if $f (- x) = f (x)$ (symmetric about the $y$ -axis) and odd if $f (- x) = - f (x)$ (symmetric about the origin).
Even $\times$ odd $=$ odd, odd $\times$ odd $=$ even, even $\times$ even $=$ even — the signs multiply.
On graphs, combine outputs vertically: add, subtract, multiply or divide the two heights at each input.

Composition of Functions & Modelling

$(f \circ g) (x) = f (g (x))$ — the inner function acts first; read the notation from the inside out.
Composition is not commutative: $f (g (x)) \neq = g (f (x))$ in general.
Domain of $f \circ g$ : all $x$ in the domain of $g$ whose output $g (x)$ also lies in the domain of $f$ .
To decompose, find the operation performed last — that becomes the outer function $f$ .
If $g$ is the inverse of $f$ , then $f (g (x)) = x$ and $g (f (x)) = x$ on the appropriate domains.
When no algebraic method exists, solve by graphing both sides or by narrowing a sign change numerically, and state the precision of the answer.

Comparing Functions and Choosing a Model

Asymptotes rule out polynomials and sinusoids; vertical asymptotes suggest rational or logarithmic, horizontal ones rational or exponential.
Exponential: domain all reals, range $y > 0$ (for $y = b^{x}$ ), horizontal asymptote. Logarithmic: domain $x > 0$ , range all reals, vertical asymptote.
Sinusoidal functions are periodic with a bounded range; polynomials have no asymptotes and end behaviour set by the leading term.
Any increasing exponential eventually exceeds any polynomial: $2^{x} > x^{3}$ for all $x > 9.94$ (approximately).
Constant nth differences signal a degree-n polynomial; a constant ratio signals an exponential.
Justify a model by naming the evidence for it and a feature that rules out the nearest alternative, then check its domain and long-run behaviour in context.

Solving Equations and Inequalities Graphically and Numerically

Solve $f (x) = g (x)$ by finding the zeros of $h (x) = f (x) - g (x)$ ; the zeros of $h$ are the $x$ -coordinates of the intersections.
$h (x) > 0$ exactly where $f (x) > g (x)$ , so the sign of $h$ between its zeros solves inequalities.
If $h$ is continuous on $[a, b]$ and $h (a)$ , $h (b)$ have opposite signs, there is at least one root in $(a, b)$ .
Bisection: test the midpoint, keep the half with the sign change; after $n$ steps the width is $\frac{b - a}{2 ^{n}}$ .
Stop when the stated accuracy is guaranteed, for example when both endpoints round to the same value.
Choose a graphing window from end behaviour so no solution is hidden off the screen, and use radians for trig equations.

Where marks get lost

Subtracting in opposite orders — computing $f (b) - f (a)$ on top but $a - b$ on the bottom — which flips the sign of the whole answer. (Average Rate of Change)
Reporting a bare number with no units, so "the rate is $6.5$ " could mean anything. (Average Rate of Change)
Treating the average rate as the value of the function, e.g. saying "the population was $6.5$ thousand" instead of "it grew by $6.5$ thousand per year". (Average Rate of Change)
Claiming an average rate of change describes every moment in the interval — a zero average rate does not mean the function stood still. (Average Rate of Change)
Dividing by the number of table rows instead of by the actual change in the input values, especially when the inputs are not spaced one unit apart. (Average Rate of Change)
Substituting $h = 0$ into the difference quotient before cancelling, which produces the meaningless form $\frac{0}{0}$ . (Instantaneous Rate of Change)
Forgetting to expand $(a + h)^{2}$ properly and writing $a^{2} + h^{2}$ , which loses the crucial $2 ah$ middle term. (Instantaneous Rate of Change)
Cancelling $h$ from only one term of the numerator instead of factoring it out of the whole numerator first. (Instantaneous Rate of Change)
Using one large value of $h$ (like $h = 1$ ) and calling the result the instantaneous rate — that is still an average rate over a wide interval. (Instantaneous Rate of Change)
Dividing by $h$ instead of $2 h$ in a centred difference, which halves the answer. (Instantaneous Rate of Change)
Concluding there is a turning point wherever the rate is zero, without checking that the rate changes sign. (Rates of Change and the Shape of a Graph)
Saying a function is decreasing because its VALUES are negative, when increasing and decreasing depend only on the sign of the rate. (Rates of Change and the Shape of a Graph)
Placing a sinusoid's fastest change at its maximum, where the rate is actually zero. (Rates of Change and the Shape of a Graph)
Confusing a value with a rate: the height at the top of a Ferris wheel is 42 m, but its rate there is 0 m/min. (Rates of Change and the Shape of a Graph)
Assuming an exponential grows at a constant rate. Its rate grows in proportion to its value. (Rates of Change and the Shape of a Graph)
Reading the domain off the simplified expression, so the restriction $x \neq = - 2$ disappears after the factor $x + 2$ cancels. (Combining Functions)
Taking the union of the two domains instead of the intersection — a combination needs values allowed by both functions, not by either one. (Combining Functions)
Forgetting the extra $g (x) \neq = 0$ condition on a quotient, especially when $g$ is a radical or a trig function that is zero at several inputs. (Combining Functions)
Dropping the bracket when subtracting, so with $f (x) = x^{2}$ and $g (x) = x - 5$ they write $(f - g) (x) = x^{2} - x - 5$ instead of the correct $x^{2} - (x - 5) = x^{2} - x + 5$ . (Combining Functions)
Declaring a function odd because it has a negative term, or even because it looks symmetric — the only valid test is substituting $- x$ and simplifying. (Combining Functions)
Composing in the wrong order — computing $g (f (x))$ when the question asked for $f (g (x))$ , or vice versa. (Composition of Functions & Modelling)
Confusing $(f g) (x)$ , which means $f (x) \cdot g (x)$ , with $(f \circ g) (x)$ , which means $f (g (x))$ . (Composition of Functions & Modelling)
Dropping brackets when substituting, so $f (g (x)) = (2 x - 3)^{2}$ is written as $2 x - 3^{2}$ . (Composition of Functions & Modelling)
Finding the domain from the simplified composite instead of from the chain, which erases restrictions the inner function imposed. (Composition of Functions & Modelling)
Grinding away at algebra on an equation like $x^{3} = 2^{x}$ that has no algebraic solution, instead of switching to a graphical or numerical method. (Composition of Functions & Modelling)
Choosing a quadratic for growth data because the first few points curve upward, without checking whether the ratios are constant. (Comparing Functions and Choosing a Model)
Assuming a polynomial with a large coefficient, such as $1000 x^{2}$ , will always beat an exponential. Any increasing exponential eventually wins. (Comparing Functions and Choosing a Model)
Confusing the domains of exponentials and logarithms: $2^{x}$ accepts every $x$ ; $lo g_{2} x$ needs $x > 0$ . (Comparing Functions and Choosing a Model)
Calling every curve with an asymptote exponential, when rational and logarithmic functions have asymptotes too. (Comparing Functions and Choosing a Model)
Justifying a model only by saying it looks right, with no differences, ratios or features cited as evidence. (Comparing Functions and Choosing a Model)
Using a window that starts at $x = 0$ and missing a negative solution, such as $x \approx - 0.767$ for $2^{x} = x^{2}$ . (Solving Equations and Inequalities Graphically and Numerically)
Keeping the wrong half in bisection: keep the half whose endpoints have OPPOSITE signs of $h$ . (Solving Equations and Inequalities Graphically and Numerically)
Applying bisection across an asymptote, where the sign changes but there is no root. (Solving Equations and Inequalities Graphically and Numerically)
Reporting a bisection midpoint to more decimal places than the final interval guarantees. (Solving Equations and Inequalities Graphically and Numerically)
Solving a trig equation such as cos x = x with the calculator in degree mode. (Solving Equations and Inequalities Graphically and Numerically)

Practice test

21 questions from across the unit. Each one shows the lesson it comes from.

Average Rate of Change
1
Which expression gives the average rate of change of $f$ from $x = a$ to $x = b$ ?
Average Rate of Change
2
For $g (x) = \frac{24}{x}$ , find the average rate of change from $x = 2$ to $x = 8$ .
Average Rate of Change
3
A function has an average rate of change of $0$ on the interval from $x = 1$ to $x = 5$ . Does that mean the function is constant on that interval? Answer yes or no.
Instantaneous Rate of Change
4
Which expression is the difference quotient for $f$ at $x = a$ ?
Instantaneous Rate of Change
5
As $h$ gets smaller and smaller, the secant line through $(a, f (a))$ and $(a + h, f (a + h))$ does what?
Instantaneous Rate of Change
6
Match each term to what it measures or means.
Secant line
Tangent line
The letter h
0/0
Rates of Change and the Shape of a Graph
7
The instantaneous rate of change of f is negative at every point of the interval 2 < x < 5. What must be true?
Rates of Change and the Shape of a Graph
8
$N (t) = 500 (2)^{t /3}$ grows at about $0.231 N$ bacteria per hour. Estimate the growth rate when $t = 6$ h, to the nearest whole number.
bacteria/h
Rates of Change and the Shape of a Graph
9
For a ball with $h (t) = - 4.9 t^{2} + 19.6 t + 2$ m, the rate of change at $t = 3$ s is $- 9.8$ m/s. What does this mean?
Combining Functions
10
From the table in this lesson, $f (2) = 3$ and $g (2) = 4$ . Find $(f + g) (2)$ .
Combining Functions
11
For the same $f (x) = x - 2$ and $g (x) = 10 - x$ , how many integer values lie in the domain of $(f + g)$ ?
Combining Functions
12
$f (x) = \frac{1}{x}$ and $g (x) = x - 5$ . State the domain of $(f + g)$ .
Composition of Functions & Modelling
13
Let $f (x) = x^{2} + 1$ and $g (x) = 2 x - 3$ . Find $f (g (2))$ .
Composition of Functions & Modelling
14
A store takes 20 dollars off a 250-dollar jacket, then charges 13% HST on the reduced price. What does the customer pay?
$
Composition of Functions & Modelling
15
A circular oil slick has radius $r (t) = 0.5 t$ km after $t$ hours, and a circle of radius $r$ has area $A (r) = π r^{2}$ . Find $A (r (6))$ , the area after 6 hours, in square kilometres, rounded to two decimal places.
Comparing Functions and Choosing a Model
16
A function has domain all real numbers, range y > 0 and a horizontal asymptote. Which family is it most likely from?
Comparing Functions and Choosing a Model
17
Order these values at $x = 20$ from smallest to largest.
1. 1 $x^{2}$
2. 2 $2^{x}$
3. 3 $x^{3}$
4. 4 $lo g_{2} x$
Comparing Functions and Choosing a Model
18
A student fits $y = 0.8 x^{2} + 2 x + 50$ to population data that turn out to grow by a constant 20 percent per year. What is the strongest objection?
Solving Equations and Inequalities Graphically and Numerically
19
Which function's zeros are the solutions of $2^{x} = x^{2}$ ?
Solving Equations and Inequalities Graphically and Numerically
20
What is the name of the numerical method that repeatedly halves an interval containing a sign change?
Solving Equations and Inequalities Graphically and Numerically
21
$h (x) = \frac{1}{x}$ has $h (- 1) < 0$ and $h (1) > 0$ . What can you conclude?

Endless drill

Fresh numbers every time, for the calculation questions in this unit.

Endless drillnew numbers every time

Generating a problem…