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a) The value of the function at a, or the image of a.

b) An asymptote is a line that approaches but does not intersect the graph of a function.

c) The largest set of values where we can evaluate the function.

d) The function’s set of images.

e) If the vertical line intersects the graph more than once, then the graph isn’t a function, because an input has more than one output.

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a) Linear function with the y intercept at 1.

b) Quadratic function, and we can check that \( g(0) = -2 \).

c) Cubic function and we can check that \( h(0) = 1 \).

For each of the graphs 1, 3, 6 and 7, we see that some points of the x-axis have more than one image, so the graphs cannot represent functions.

5 is the graph of the function \(𝑥 ↦ \sqrt{x+1}\)

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a)

(a)	Algebraically	Graphically
Domain	We need the number under the square root to be ≥ 0, so 4𝑥 − 8 ≥ 0. This means the domain is {𝑥\|𝑥 ∈ R, 𝑥 ≥ 2}.	The smallest value of x with an image is 2,so the domain is 𝑥 ≥ 2.
Range	\(\sqrt{4x-8}\) is always ≥ 0, so \(\sqrt{4x-8} - 2\) is always ≥ −2. The range is {𝑦\|𝑦 ∈ R, 𝑦 ≥ 2}.	The values for y start at -2, so the range is 𝑦 ≥ −2.

b)

(b)	Algebraically	Graphically
Domain	We can’t divide by zero, so we need 𝑥 ≠ 3. The domain is {𝑥\|𝑥 ∈ R,𝑥 ≠ 3}.	There is an asymptote for x=3, where the function has no value. So we know the domain is all of R where 𝑥 ≠ 3.
Range	The only number not in the range of \(\frac{1}{3-x}\)is 0. Since we add 2 to all values, the range of the function is all of R except x = 2. The range is: {𝑦\|𝑦 ∈ R, 𝑦 ≠ 2}.	All values of y are reached except y=2 where there is another asymptote. So the range is R where 𝑦 ≠ 2.

c)

(c)	Algebraically	Graphically
Domain	We can’t divide by zero, so we must have 𝑥³ − 1 ≠ 0, which means 𝑥 ≠ 1. The domain is {𝑥\|𝑥 ∈ R, 𝑥 ≠ 1}.	It’s not clear here if there is an asymptote at x = 1 or If the graph has a “peak” and goes back down, which is why it’s important to also do the algebraic approach. From the graph, the domain can be R or R with 𝑥 ≠ 1.
Range	The denominator is always positive, so \(\frac{3}{{(x^3-1)}^2}\) hits all positive values except 0. This means the range is {𝑦\|𝑦 ∈ R, 𝑦 > 0}.	All values strictly above 0 appear to be reached, so the range is 𝑦 > 0.

d)

(d)	Algebraically	Graphically
Domain	We can’t divide by zero, so we must have 𝑥² − 1 ≠ 0, which means 𝑥 ≠ 1and 𝑥 ≠ −1.The domain is {𝑥\|𝑥 ∈ R, 𝑥 ∉ {−1,1}}.	There are two asymptotes for x = -1 and x = 1, so the domain is all of R where 𝑥 ≠ −1 and 𝑥 ≠ 1.
Range	Like before, denominator always positive and fraction always non null, so the range is {𝑦\|𝑦 ∈ R, 𝑦 > 0}.	All values strictly above 0 appear to be reached again, so again the range is 𝑦 > 0.

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a) Each point on the x-axis has at most one value associated to it, so this is a function.

b) The function is defined everywhere except for x=5, where there is an asymptote.

c) Graphically, we can see that the function only has positive values. It also looks like the graph never reaches the x-axis, so the values are strictly positive.

d) We can read them all graphically.

e) 4 and 6 for example have the same image, so this is not a one-to-one function.

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a)

\[ f(2) = 1 + 2^2 = 5\]

\[ g(2) = \frac{1}{1+2} = \frac{1}{3}\]

b) The domain of \( f \) is \( \mathbb{R} \), but \( g \) is not defined for \( x = -1 \) (because then we would divide by zero), so we have to exclude it from the domain of \( f + g \).

c) The range of \( f(x) \) is \( y > 1 \), but \( g(x) \) reaches values \( \leq 1 \), and the range of \( g(x) \) is \( y \neq 0 \), but \( f(x) \) reaches 0, so the total range is \( \mathbb{R} \).

We can also see this graphically:

d)

\[ (f+g)(2) = f(2) + g(2) = 5 + \frac{1}{3} = \frac{16}{3}\]

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a) The domain is defined if: \(\sqrt{4-x} \neq 0\) .

\(\sqrt{4-x} \neq 0\) is defined for \( x \leq 4 \), only reaching 0 at \( x = 4 \). So the domain of \( f \) is \( \{ x \in \mathbb{R} \mid x < 4 \} \).

As for the range, \(\frac{4}{\sqrt{4-x}}\) is always greater than 0, and we have to subtract 2 (as the function subtracts 2 from the fraction). Therefore, the range is for \( y > -2 \).

b) The asymptotes can be clearly seen by looking at the graph of the function or analyzing the domain and range from part (a). Therefore, the vertical asymptote is at \( x = 4 \), and the horizontal asymptote is at \( y = -2 \).

c) The inverse can be easiest found by first flipping the order of \( y \) and \( x \) in the original function, and then solving for \( y \):

\[ y = \frac{4}{\sqrt{4-x}} - 2 \]

\[ x = \frac{4}{\sqrt{4-y}} - 2 \]

\[ x + 2 = \frac{4}{\sqrt{4-y}} \]

\[ (x + 2) \sqrt{4-y} = 4 \]

\[ \sqrt{4-y} = \frac{4}{x+2} \]

\[ 4-y = \left(\frac{4}{x+2}\right)^2 \]

\[ y = 4 - \left(\frac{4}{x+2}\right)^2 \]

d) The domain and range of the inverse can be easily found by applying the rule stating that the domain of \( f(x) \) is equal to the range of \( f^{-1}(x) \) and the range of \( f(x) \) is equal to the domain of \( f^{-1}(x) \). By applying this rule, we can easily find that the domain of the inverse is \( x > -2 \), and the range is \( y < 4 \).

e) and f) Both of these subquestions can be answered using the graph below:

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a) We use one of the easily read coordinates on the graph, for example (0,2):

\[ f(2) = \frac{(2+k)^2}{2}=0 \implies (2+k)^2 = 0 \implies k = -2\]

As for the range, \(\frac{4}{\sqrt{4-x}}\) is always greater than 0, and we have to subtract 2 (as the function subtracts 2 from the fraction). Therefore, the range is for \( y > -2 \).

b) To answer this question we use the fact that the domain of \( f(x) \) is equal to the range of its inverse, and vice versa. Therefore, the first point we're looking for, \( f^{-1}(2) \), is at the point where \( y = 2 \) for the original function. By looking at the graph, we can see that \( y = 2 \) occurs at two points, \( x = 0 \) and \( x = 4 \).

Similarly, \( f^{-1}(0) \) will happen when \( y = 0 \) for the original function. We can clearly see that for the original function, \( y = 0 \) at \( x = 2 \).

c) We can see that the graph is symmetrical with respect to \( x = 2 \). Alternatively, we can rewrite \( f(x) \) as \( f(x) = \frac{1}{2}x^2 - 2x + 2 \) , which is a quadratic function with axis of symmetry at:

\[ x = - \frac{b}{2a} = \frac{2}{2 \cdot \frac{1}{2}} = 2 \]

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Detailed Answer

a) It can be clearly seen from the function's graph that the vertical asymptote is at \( x = 1 \), and the horizontal asymptote is at \( y = 2 \).

b) Since we don't know the equation of the function, we can't directly plug points into the inverse formula. However, knowing the relationship between the domain and range of inverse functions, we can find those points by looking at the graph. Therefore, taking the first point as an example, \( f^{-1}(0) \) is actually the point for which \( y = 0 \) for the function \( f(x) \). As we can see, the x-coordinate for this point is \( x = 2 \).

For the second point, \( f^{-1}(2) \) means we are looking for the point when \( y = 2 \) in \( f(x) \). As we can see, \( y = 2 \) is undefined for the original function since this is the value of one of the asymptotes.

The third point, \( f^{-1}(-2) \), means we are looking for when \( y = -2 \) for the original function. At \( y = -2 \), we can see that \( x = 0.5 \), which means that \( f^{-1}(-2) = 0.5 \).

Lastly, using the same approach, when \( y = 4 \) for the original function, then \( x = 2 \), meaning that \( f^{-1}(4) = 2 \).

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a)

\[ f(2) = 0.5 * 2 + 2 = 3\]

\[ f(4) = 0.5 * 4 + 2 = 4\]

b) This can be easily done in GDC resulting in the following graph.

c) This question can either be answered by looking at the original graph of \( f(x) \) or by calculating the inverse function:

\[ y = \frac{1}{2}x + 2\]

\[ x = \frac{1}{2}y + 2\]

\[ x - 2 = \frac{1}{2}y\]

\[ y = 2x - 4\]

Then, by plugging those two points into this inverse function we get that:

\[ f^{-1}(2) = 2*2 - 4 = 0\]

\[ f^{-1}(3.5) = 2*3.5 - 4 = 3\]

d) This question can be solved in multiple ways.

Firstly, from the previous question, we know the coordinates of two points of this line, and we can draw it.

Alternatively, we can draw the line \( y=x \), and draw the symmetry of \( f \) with regards to it.

We can also place the point with coordinates (3,2) and draw the line going through it and (4,4).

Or you can just put it into GDC 😉

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a) To find the time taken for Ted to run the flat section, we use the formula: \( t = \frac{d}{v} = \frac{6}{12} = 0.5 \). The time for the incline is given by \( \frac{10}{x} \) . The average speed can be calculated by dividing the total distance by the total time:

\[ Av = \frac{6 + 10}{0.5 + \frac{10}{x}} = \frac{32x}{x + 20} \]

b) To find the inverse function, we rearrange \( x \) and \( y \) and solve for \( y \):

\[ x = \frac{32y}{y + 20} \]

\[ y = \frac{-20x}{x - 32} \]

\[ Av^{-1}(x) = \frac{-20x}{x - 32} \]

c)

\[ Av^{-1}(10) = \frac{-20(10)}{10 - 32} = 9.09 \; \text{km/h} \]

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Detailed Answer

a) To find the range of the function, we can graph it:

We can see that the range is: \( (0,\infty) \cup (-\infty,-0.346] \)

b) To find \( f^{-1}(-0.5) \), we look for the x-value where \( f(x) = -0.5 \). From the graph, we find that:

\( x = 1 \) .

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Detailed Answer

a) \( f(-12) = 6 \) and \( f(8) = 2 \). So the range is \( f(x) \ge 6 \) and \( f(x) \le 2 \).

b) To find \( f^{-1}(x) \), we switch \( x \) and \( y \) in the original function and solve for \( y \):

\[ x = 5 - \frac{15}{y-3} \]

Rearranging gives:

\[ \frac{15}{y-3} = 5 - x \implies 15 = (5-x)(y-3) \]

\[ 15 = (5-x)y - 3(5-x) \]

\[ 15 + 15 - 3x = (5-x)y \]

\[ y = \frac{30 - 3x}{5 - x} \]

c) The range of the inverse is the domain of \( f(x) \), thus \( -12 \le f^{-1}(x) < 8 \), \( f^{-1}(x) \neq 3 \).

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a) To find the inverse, we exchange \( x \) and \( y \) and solve for \( y \):

\[ x = \ln\left(\frac{1}{y+5}\right) \]

\[ e^x = \frac{1}{y+5} \]

\[ y+5 = \frac{1}{e^x} \]

\[ y = \frac{1}{e^x} - 5 \]

\[ f^{-1}(x) = \frac{1}{e^x} - 5 \]

b) To solve \( f(x) = f^{-1}(x) \), we can graph both functions and find their intercept:

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Properties of Functions

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Question 1

[Maximum mark: 6]

Answer those general questions on functions:

a) If a function \( f \) is evaluated at a, what is a name for \( f(a) \)?

b) What is an asymptote of a function?

c) Define the domain of a function.

d) Define the range of a function.

e) How can the vertical line test help us recognize a function?

Answers and Explanations

Question 2

[Maximum mark: 8]

Match each function below to its graph. Say which graphs don’t show a function.

a) \(f:x↦1+x\)

b) \(g:x↦x^2-2\)

c) \(h:x↦\frac{1}{2}x^3 + \frac{1}{3}x^2 + x + 1\)

Answers and Explanations

Question 3

[Maximum mark: 10]

Consider a function \( f(x) \) displayed on the graph below

a) Give the equations of two asymptotes of \( f(x) \).

b) Find the values of: \( f^{-1}(0) \), \( f^{-1}(2) \), \( f^{-1}(-2) \), \( f^{-1}(4) \)

Answers and Explanations

Question 4

[Maximum mark: 9]

Consider a function displayed on the graph below:

a) Graphically, check that this is indeed a function.

b) What is the domain of \( f \)?

c) What is the range of \( f \)?

d) Find the values of \( f(5) \), \( f(4) \), and \( f(6) \).

e) Is this a one-to-one function?

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Question 5

[Maximum mark: 12]

Give the domain and range of the following functions, algebraically and graphically.

a) \(f(x) = \sqrt{4x-8} - 2\)

b) \(f(x) = \frac{1}{3-x} + 2\)

c) \(f(x) = \frac{3}{(x^3-1)^2}\)

d) \(f(x) = \frac{3}{(x^2-1)^2}\)

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Question 6

[Maximum mark: 16]

Consider a function \(f(x) = \frac{4}{\sqrt{4-x}} - 2\)

a) State the domain and range of \( f(x) \).

The graph of this function looks as follows:

b) State the equation of two asymptotes.

c) Find the equation for the inverse of \( f(x) \).

d) State the domain and range of \( f^{-1}(x) \).

e) Find graphically the interesction coordinates of \( f(x) \) and \( f^{-1}(x) \).

f) Sketch the graphs of \( f(x) \), \( y = x \), and \( f^{-1}(x) \) on the same diagram.

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Question 7

[Maximum mark: 10]

Consider a function \(f(x) = 0.5x + 2\)

a) Find the values of \( f(2) \) and \( f(4) \).

b) Draw the graph of \( f(x) \).

c) Find \( f^{-1}(2) \) and \( f^{-1}(3.5) \)

d) Draw the graph of \( f^{-1}(x) \) on the same diagram as before.

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Question 8