Higher Maths
Functions and Graphs

The denominator of a fraction cannot equal zero. So the solutions of this equation cannot be in the domain:

$$\begin{array}{rcl}{x}^2-4x-5& =& 0\\ (x+1)(x-5)& =& 0\end{array}$$

$$x=-1\mathsf{\text{or}}x=5$$

The only restriction on the domain of $g$ is that we cannot take the square root of a negative number. So:

$$\begin{array}{rcl}{x}^2+x-12& ⩾& 0\\ (x+4)(x-3)& ⩾& 0\end{array}$$

$$x⩽-4\mathsf{\text{or}}x⩾3$$

So the maximal domain of $g$ is $\{x:x\in \mathbb{R},x⩽-4,x⩾3\}.$

From Nat 5, you should know that this is the basic $y=cosx$ graph translated up $1$.

The domain is restricted to the real values between $\frac{\pi }{2}$ radians $(={90}^{\circ })$ and $\frac{3\pi }{2}$ radians $(={270}^{\circ }),$ inclusive.

A quick, rough sketch of the graph should let you confirm that the range of $h$ is $\{y:y\in \mathbb{R},0⩽y⩽1\}.$

(a)  Note that $f(g(x))$ means that $g(x)$ is the input for $f$. So:

$$\begin{array}{rcl}f(g(x))& =& f(3x^2-5)\\ & =& 1-2(3x^2-5)\\ & =& 1-6x^2+10\\ & =& 11-6x^2\end{array}$$

(b)  Similarly, $g(f(x))$ means that $f(x)$ is the input for $g$. So:

$$\begin{array}{rcl}g(f(x))& =& g(1-2x)\\ & =& 3(1-2x{)}^2-5\\ & =& 3(1-4x+4x^2)-5\\ & =& 3-12x+12x^2-5\\ & =& 12x^2-12x-2\end{array}$$

Note that $f^2(x)$ means $f(f(x)),$ so we need to feed $f(x)$ back into $f$ again.

$$\begin{array}{rcl}{f}^2(x)& =& f(f(x))\\ & =& f\left(\frac{x-1}{x+1}\right)\\ & =& \frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}\\ & =& \frac{(x-1)-(x+1)}{(x-1)+(x+1)}\\ & =& \frac{x-1-x-1}{x-1+x+1}\\ & =& \frac{-2}{2x}\\ & =& -\frac{1}{x}\end{array}$$

(a)  The SQA will accept two different methods of finding an inverse function.

Method 1:

Use the definition of $f^{-1}$ as the function for which $f(f^{-1}(x))=x,$ and rearrange:

$$\begin{array}{rcl}f(f^{-1}(x))& =& x\\ 3-2(f^{-1}(x))& =& x\\ -2f^{-1}(x)& =& x-3\\ 2f^{-1}(x)& =& 3-x\\ f^{-1}(x)& =& \frac{3-x}{2}\end{array}$$

Method 2:

This method uses the fact that if $y=f(x),$ then $x=f^{-1}(y).$

So we write $f(x)$ as $y$ and rearrange to make $x$ the subject:

$$\begin{array}{rcl}f(x)& =& 3-2x\\ y& =& 3-2x\\ 2x& =& 3-y\\ x& =& \frac{3-y}{2}\\ f^{-1}(y)& =& \frac{3-y}{2}\\ \therefore f^{-1}(x)& =& \frac{3-x}{2}\end{array}$$

(b)  This would only be a one-mark question. We are not told the rule for the function $g$ so it relies upon us understanding that an inverse function performs the opposite operation to the original function. So if $g(4)=5,$ then $g^{-1}(5)=4.$ It's as simple as that!

(c)  This would also only be worth one mark. Again, we need to understand that substituting any value into a function and then feeding the output into its inverse function takes us back to the original value. So without any need for working, $g(g^{-1}(x))=x.$ Similarly, if we had been asked to write down an expression for $g^{-1}(g(x)),$ that would also just be $x.$

(a)  $y=f(x-3)$ translates the graph of $y=f(x)$ to the right by $3$ units. So the turning points become $(3,2)$ and $(6,-1).$

(b)  $y=2f(x)$ 'stretches' the graph $y=f(x)$ vertically. Each $y$-coordinate is multiplied by $2.$ So the turning points are $(0,4)$ and $(3,-2).$

(c)  $y=-4f(x+2)$ involves both a vertical and a horizontal transformation. The $y$-coordinates are multiplied by $-4,$ which has the effect of 'stretching' the graph to $4$ times its original height and also reflecting it in the $x$-axis. The graph is also translated $2$ units to the left. So the turning points are $(-2,-8)$ and $(1,4).$

(d)  $y=f(3x)-1$ also involves both a vertical and a horizontal transformation. The $y$-coordinates are reduced by $1$ and the $x$-coordinates are divided by $3.$ So the turning points are $(0,1)$ and $(1,-2).$

(e)  $y=\frac{1}{2}f(x)+4$ involves two vertical transformations. First, each $y$-coordinate is multiplied by $\frac{1}{2}$ (that is, divided by $2$) and then $4$ is added, which translates the graph up $4$ units. So the turning points are $(0,5)$ and $(3,\frac{7}{2}).$

(a)  Note that $f(g(x))$ means that $g(x)$ is the input for $f$. So:

$$\begin{array}{rcl}f(g(x))& =& f(5-x)\\ & =& \frac{1}{\sqrt{5-x}}\end{array}$$

(b)  We cannot square root a negative number or divide by zero, so $f(g(x)$ is undefined when:

$$\begin{array}{rcl}5-x& \le & 0\\ 5& \le & x\\ x& \ge & 5\end{array}$$

(a)  The SQA will accept two different methods of finding an inverse function.

Method 1:

Use the definition of $f^{-1}$ as the function for which $f(f^{-1}(x))=x,$ and rearrange:

$$\begin{array}{rcl}f(f^{-1}(x))& =& x\\ \sqrt[3]{f^{-1}(x)}+8& =& x\\ \sqrt[3]{f^{-1}(x)}& =& x-8\\ f^{-1}(x)& =& (x-8{)}^3\end{array}$$

Method 2:

This method uses the fact that if $y=f(x),$ then $x=f^{-1}(y).$

So we write $f(x)$ as $y$ and rearrange to make $x$ the subject:

$$\begin{array}{rcl}f(x)& =& \sqrt[3]{x}+8\\ y& =& \sqrt[3]{x}+8\\ y-8& =& \sqrt[3]{x}\\ (y-8{)}^3& =& x\\ x& =& (y-8{)}^3\\ f^{-1}(y)& =& (y-8{)}^3\\ f^{-1}(x)& =& (x-8{)}^3\end{array}$$

(b)  This was a one-mark question, so it was fine to write down the answer without any working.

Key fact: the domain of $f^{-1}$ equals the range of $f.$

As $f$ is an increasing function, all we have to do here is to evaluate $f$ at the lowest and highest values of its domain.

$f(1)=\sqrt[3]{1}+8=1+8=9$
$f(1000)=\sqrt[3]{1000}+8=10+8=18$

So the domain of $f^{-1}$ is $9⩽x⩽18,x\in \mathbb{R}.$

The SQA will accept two different methods of finding an inverse function.

Method 1:

Use the definition of $h^{-1}$ as the function for which $h(h^{-1}(x))=x,$ and rearrange:

$$\begin{array}{rcl}h(h^{-1}(x))& =& x\\ 4+\frac{1}{3}{h}^{-1}(x)& =& x\\ \frac{1}{3}{h}^{-1}(x)& =& x-4\\ h^{-1}(x)& =& 3(x-4)\end{array}$$

Method 2:

This method uses the fact that if $y=h(x),$ then $x=h^{-1}(y).$

So we write $h(x)$ as $y$ and rearrange to make $x$ the subject:

$$\begin{array}{rcl}h(x)& =& 4+\frac{1}{3}x\\ y& =& 4+\frac{1}{3}x\\ y-4& =& \frac{1}{3}x\\ 3(y-4)& =& x\\ x& =& 3(y-4)\\ h^{-1}(y)& =& 3(y-4)\\ h^{-1}(x)& =& 3(x-4)\end{array}$$

The SQA will accept two different methods of finding an inverse function.

Method 1:

Use the definition of $f^{-1}$ as the function for which $f(f^{-1}(x))=x,$ and rearrange:

$$\begin{array}{rcl}f(f^{-1}(x))& =& x\\ \frac{2}{f^{-1}(x)}+3& =& x\\ \frac{2}{f^{-1}(x)}& =& x-3\\ 2& =& (x-3)f^{-1}(x)\\ f^{-1}(x)& =& \frac{2}{x-3}\end{array}$$

Method 2:

This method uses the fact that if $y=f(x),$ then $x=f^{-1}(y).$

So we write $f(x)$ as $y$ and rearrange to make $x$ the subject:

$$\begin{array}{rcl}f(x)& =& \frac{2}{x}+3\\ y& =& \frac{2}{x}+3\\ y-3& =& \frac{2}{x}\\ (y-3)x& =& 2\\ x& =& \frac{2}{y-3}\\ f^{-1}(y)& =& \frac{2}{y-3}\\ f^{-1}(x)& =& \frac{2}{x-3}\end{array}$$

The SQA will accept two different methods of finding an inverse function.

Method 1:

Use the definition of $h^{-1}$ as the function for which $h(h^{-1}(x))=x,$ and rearrange:

$$\begin{array}{rcl}h(h^{-1}(x))& =& x\\ 2(h^{-1}(x){)}^3-7& =& x\\ 2(h^{-1}(x){)}^3& =& x+7\\ (h^{-1}(x){)}^3& =& \frac{x+7}{2}\\ h^{-1}(x)& =& \sqrt[3]{\frac{x+7}{2}}\end{array}$$

Method 2:

This method uses the fact that if $y=h(x),$ then $x=h^{-1}(y).$

So we write $h(x)$ as $y$ and rearrange to make $x$ the subject:

$$\begin{array}{c}h(x)=2x^3-7\\ y=2x^3-7\\ y+7=2x^3\\ \frac{y+7}{2}=x^3\\ \sqrt[3]{\frac{y+7}{2}}=x\\ x=\sqrt[3]{\frac{y+7}{2}}\\ h^{-1}(y)=\sqrt[3]{\frac{y+7}{2}}\\ h^{-1}(x)=\sqrt[3]{\frac{x+7}{2}}\end{array}$$