2.7 Summary

\begin{align*} \text{Gradient: } m &= \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ &= \frac{-6 - 0}{0 - (-3)} \\ &= \frac{-6}{3} \\ &= -2\\ \therefore h(x) &= -2x - 6 \end{align*}

\begin{align*} \text{Let } y &= -2x - 6 \\ \text{Inverse: } x &= -2y - 6 \\ x + 6 &= -2y \\ -\frac{1}{2} (x + 6) &= y \\ y &= -\frac{x}{2} - 3 \\ \therefore h^{-1}(x) &= -\frac{x}{2} - 3 \end{align*}

\begin{align*} -\frac{x}{2} - 3 &= -2x - 6 \\ -x - 6 &= -4x - 12 \\ -x + 4x &= -12 + 6 \\ 3x &= -6 \\ \therefore x &= -2 \\ \text{If } x=-2, \quad y &= -2(-2) - 6 \\ y &= -2(-2) - 6 \\ &= -2 \end{align*}

This gives the point \(S(-2;-2)\)

The value of the \(x\)-coordinate and the \(y\)-coordinate will always be the same since the point lies on the line \(y = x\).

\begin{align*} \text{Let } y &= 2x + 4 \\ \text{Inverse: } x &= 2y + 4 \\ x - 4 &= 2y \\ \frac{1}{2}x - 2 &= y \\ \therefore f(x) &= \frac{1}{2}x - 2 \end{align*}

Increasing. As \(x\) increases, the function value increases. Alternative reason: gradient is positive, therefore function is increasing.

The domain is: \(\{x: x \in \mathbb{R} \}\) and the range is: \(\{y: y \geq 0, y \in \mathbb{R} \}\).

\begin{align*} \text{Let } y &= 2x^{2} \\ \text{Inverse: } x &= 2y^{2} \\ \frac{1}{2}x &= y^{2} \\ \pm \sqrt{ \frac{1}{2}x } &= y \\ \therefore y &= \pm \sqrt{ \frac{x}{2} } \qquad (x \ge 0) \end{align*}

Domain: \(\{x: x \geq 0, x \in \mathbb{R} \}\), Range: \(\{y: y \in \mathbb{R} \}\).

\begin{align*} f(x) & = \left( \frac{1}{4} \right)^{x} \\ \text{Substitute } \left( -\frac{1}{2}; 2 \right): \quad f\left(-\frac{1}{2}\right) & = \left( \frac{1}{4} \right)^{-\frac{1}{2}}\\ & = 4^{\frac{1}{2}} \\ & = 2 \end{align*}

Yes, the point \(\left( -\frac{1}{2}; 2 \right)\) does lie on \(f\).

\begin{align*} f: \quad y & = \left( \frac{1}{4} \right)^{x} \\ f^{-1}: \quad x & = \left( \frac{1}{4} \right)^{y} \\ \therefore y & = \log_{\frac{1}{4}}{x}\\ \text{or } & \\ f: \quad y & = \left( 4 \right)^{-x} \\ f^{-1}: \quad x & = \left(4 \right)^{-y} \\ -y & = \log_{4}{x}\\ \therefore y & = - \log_{4}{x} \end{align*}

\(y = \log_{\frac{1}{4}}{x}\) or \(y = - \log_{4}{x}\)

\(P = \frac{1}{2}\), since the point lies on the line \(y = x\).

Vertical asymptote: \(x = -2\)

\begin{align*} \text{Let: } \quad y & = 3^{x} \\ \text{Inverse: } x & = 3^{y} \\ y & = \log_{3}{x} \\ \therefore h^{-1}(x) & = \log_{3}{x} \end{align*}

Domain: \(\{x: x > 0, x \in \mathbb{R} \}\) and range: \(\{y: y \in \mathbb{R} \}\).

\(0 < x < 1\)

\begin{align*} f(x) & = 2^{x} \\ f(-1) & = 2^{-1} \\ & = \frac{1}{2}\\ g(x) & = x^{2} \\ g(-1) & = (-1)^{2} \\ & = 1 \\ \therefore f(-1) & \ne g(-1) \end{align*}

The graphs do not intersect at \(x = -1\).

Two

\begin{align*} y &= \log_{b}{x} \\ 2 &= \log_{b}{9} \\ \therefore 9 &= b^{2} \\ \therefore 3^2 &= b^{2} \\ \therefore b &= 3 \end{align*}

\begin{align*} y &= \log_{3}{x} \\ -1 &= \log_{3}{a} \\ \therefore 3^{-1} &= a \\ \therefore a &= \frac{1}{3} \end{align*}

\(y = \log_{3}{x} + 2\)

\(y = \log_{3}{(x - 1)}\)

\begin{align*} A &= P \left( 1 - i \right)^{n} \\ \frac{1}{2} &= \left( 1 - \frac{7}{100}\right)^{n} \\ \text{0,5} &= \left( \text{0,93} \right)^{n} \\ \log{\text{0,5}} &= \log{\left( \text{0,93} \right)^{n}} \\ \log{\text{0,5}} &= n \log{\left( \text{0,93} \right)} \\ \therefore n &= \frac{\log{\text{0,5}} }{\log{\left( \text{0,93} \right)}} \\ &= \text{9,55} \ldots \end{align*}

It will take less than \(\text{10}\) years for the current rhino population to halve in size.

Graph C

Currently \((n = 0)\) the rhino population is \(P\). After \(\text{9,6}\) years, it will have halved, \(\frac{P}{2}\). Note: the line in the graph indicates the trend, rhino population numbers are discrete values and should be plotted points.

\(\text{100} \qquad \text{100} \times 2 \qquad \text{100} \times 2^{2} \qquad \text{100} \times 2^{3}\)

This is a geometric sequence: \(r = 2\) and \(a = 100\).

Therefore \(T_{n} = 100 \times 2^{n-1}\).

\begin{align*} T_{n} &= 100 \times 2^{n-1} \\ T_{12} &= 100 \times 2^{11} \\ &= \text{204 800} \end{align*}

\(\text{204 800}\) retweets.

\begin{align*} 200 \times 10^{6} &= 2 \times 10^{8} \\ S_{n} &= \frac{a(r^{n} - 1)}{r - 1} \\ \therefore \frac{100(2^{n} - 1)}{2 - 1} & > \text{2} \times 10^{\text{8}} \\ 2^{n} & > \frac{\text{2} \times 10^{\text{8}} }{100} + 1 \\ 2^{n} & > \text{2 000 001} \\ {n} & > \log_{2}{\text{2 000 001}} \qquad (\text{use definition}) \\ n & > \frac{ \log{\text{2 000 001}}}{\log2} \qquad (\text{change of base}) \\ n & > \text{20,9} \ldots \quad 5 \text{ minute periods} \end{align*}

Therefore, \(\frac{21}{12} = \text{1,75}\) hours.

\begin{align*} g(x) & = -1 + \sqrt{x} \quad (x \ge 0) \\ \text{Let } y & = -1 + \sqrt{x} \\ \text{Inverse: } x & = -1 + \sqrt{y} \quad (y \ge 0) \\\\ \sqrt{y} & = x + 1 \qquad (x \ge -1)\\ \therefore y & = (x + 1)^{2} \\ & \\ \therefore g^{-1}(x) & = (x + 1)^{2} \qquad (x \ge -1) \end{align*}

Yes. It passes vertical line test.

Domain: \(\{x: x \ge -1, x \in \mathbb{R} \}\), Range: \(\{y: y \geq 0,y \in \mathbb{R} \}\).

First use the information provided in the graph of the inverse:

\begin{align*} \text{Turning point: } & (3;1) \\ x-\text{intercept: } & (1;0) \end{align*}

To get the turning point and intercepts of the function, we invert the given coordinates. Now we can use those coordinates to find the equation of the function:

Now find the equation of the function:

\begin{align*} \text{Turning point: } & (1;3) \\ x-\text{intercept: } & (0;1) \\ y & = a(x-p)^{2} + q \\ y & = a(x-1)^{2} + 3 \\ \text{Substitute } (0;1) \quad 1 & = a(0-1)^{2} + 3\\ a & = -\text{2}\\ \therefore y & = -2(x-1)^{2} + 3 \end{align*}

Maximum value at \((1;3)\)

Domain: \(\{x: x \in \mathbb{R} \}\) and range: \(\{y: y \leq 3,y \in \mathbb{R} \}\), Axis of symmetry: \(x =1\).

\begin{align*} k(x) & = 2x^{2} + 1 \\ \text{Substitute } (q;3) \quad 3 & = 2(q)^{2} + 1 \\ 2 & = 2(q)^{2} \\ 1 & = q^{2} \\ \therefore q & = \pm 1 \end{align*}

This gives the points \((-1;3)\) and \((1;3)\).

\begin{align*} k: \quad y & = 2x^{2} + 1 \\ \text{Inverse: } \quad x & = 2y^{2} + 1 \\ 2y^{2} & = x - 1 \\ y^{2} & = \frac{x-1}{2} \\ y & = \pm \sqrt{\frac{x-1}{2}} \qquad (x \ge 1) \end{align*}

\((3;1)\)

\begin{align*} q & = -2 \\ y & = ax^{2} - 2 \\ \text{Substitute } (-2;-6) \quad -6 & = a(-2)^{2} - 2 \\ -6 + 2 & = 4a \\ -4 & = 4a \\ -1 & = a \\ \therefore f(x) & = -x^{2} - 2 \end{align*}

\begin{align*} \text{Let } y & = -x^{2} - 2 \qquad (y \leq -2) \\ \text{Interchange } x \text{ and } y: \quad x & = -y^{2} - 2 \qquad (x \leq -2) \\ x + 2 & = -y^{2} \\ -x - 2 & = y^{2} \\ y & = \pm \sqrt{-x - 2} \qquad (x \leq -2) \end{align*}

The inverse is a not a function. The turning point of the inverse is \((-2;0)\) and \(x\)-intercept is \((-2;0)\).

\[\text{Inverse}: \qquad \text{domain } \{x: x \le -2, x \in \mathbb{R} \} \quad \text{range } \{y: y \in \mathbb{R} \}\]

\begin{align*} \text{Let } y & = x^{2} - 9 \qquad (y \geq -9) \\ \text{Interchange } x \text{ and } y: \quad x & = y^{2} - 9 \qquad (x \geq -9) \\ x + 9 & = y^{2} \\ y & = \pm \sqrt{x + 9} \qquad (x \geq -9) \end{align*}

Determine the intercepts:

\begin{align*} \text{Let } x = 0: \quad y & = (0)^{2} - 9 \\ & = - 9 \\ \text{Let } y = 0: \quad 0 & = x^{2} - 9 \\ x^2 & = 9 \\ \therefore x & = \pm 3 \end{align*}

The intercepts are \((0;-9)\) and \((-3;0),(3;0)\).

No. The inverse does not pass the vertical line test. It is a one-to-many relation.

If the domain of \(H\) is restricted to \(\{x: x \le 0 \}\), then the inverse is \(H^{-1}(x) = - \sqrt{x^{2} + 9} \quad (x \ge -9, y \le 0)\).

The graph of \(H^{-1}\) cuts a vertical line only once at any one time and therefore passes the vertical line test.