Higher Maths
Functions and Graphs

The denominator of a fraction cannot equal zero. However, this occurs when:

$$ \begin{gather} x^2-4x-5=0 \\[6pt] (x+1)(x-5)=0 \\[6pt] x=-1\:\ \small\textsf{or}\normalsize\:\:x=5 \end{gather} $$

So \(x=-1\) and \(x=5\) must be excluded from the domain of \(f\small.\)

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

$$ \begin{gather} x^2+x-12 \geqslant 0 \\[6pt] (x+4)(x-3) \geqslant 0 \\[6pt] x\leqslant -4 \:\ \small\textsf{or}\normalsize\:\:x\geqslant 3 \\[6pt] \end{gather} $$

So the maximal domain of \(g\) is \(\{x : x\in \mathbb R , x\leqslant -4, x\geqslant 3\}\small.\)

From Nat 5, you should know that this is the basic \(y=\text{cos}\,x\) graph translated up \(1\small.\)

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

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

(a)  Note that \(f\left(g(x)\right)\) means that \(g(x)\) is the input for \(f\small.\) So:

$$ \begin{eqnarray} f\left(g(x)\right) &=& f\left(3x^2-5\right) \\[6pt] &=& 1-2(3x^2-5) \\[6pt] &=& 1-6x^2+10 \\[6pt] &=& 11-6x^2 \end{eqnarray} $$

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

$$ \begin{eqnarray} g\left(f(x)\right) &=& g\left(1-2x\right) \\[6pt] &=& 3(1-2x)^2-5 \\[6pt] &=& 3(1-4x+4x^2)-5 \\[6pt] &=& 3-12x+12x^2-5 \\[6pt] &=& 12x^2-12x-2 \end{eqnarray} $$

Note that \(f^{2}(x)\) means \(f\left(f(x)\right),\) so we need to feed \(f(x)\) back into \(f\) again.

$$ \begin{eqnarray} f^{2}(x) &=& f\left(f(x)\right) \\[10pt] &=& f\left(\small\frac{x-1}{x+1}\right)\normalsize \\[10pt] &=& \frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1} \\[10pt] &=& \small\frac{(x-1)-(x+1)}{(x-1)+(x+1)}\normalsize \\[10pt] &=& \small\frac{x-1-x-1}{x-1+x+1}\normalsize \\[10pt] &=& \small\frac{-2}{2x}\normalsize \\[10pt] &=& -\!\small\frac{1}{x}\normalsize \end{eqnarray} $$

(a)  Qualifications Scotland will accept two different methods of finding an inverse function.

Method 1:

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

$$ \begin{eqnarray} f\left(f^{-1}(x)\right) &=& x \\[6pt] 3-2\left(f^{-1}(x)\right) &=& x \\[6pt] -2\,f^{-1}(x) &=& x-3 \\[6pt] 2\,f^{-1}(x) &=& 3-x \\[6pt] f^{-1}(x) &=& \small\frac{3-x}{2}\normalsize \end{eqnarray} $$

Method 2:

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

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

$$ \begin{eqnarray} f(x) &=& 3-2x \\[6pt] y &=& 3-2x \\[6pt] 2x &=& 3-y \\[6pt] x &=& \small\frac{3-y}{2}\normalsize \\[6pt] f^{-1}(y) &=& \small\frac{3-y}{2}\normalsize \\[6pt] \therefore\,f^{-1}(x) &=& \small\frac{3-x}{2}\normalsize \end{eqnarray} $$

(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 \((6,\,-1)\small.\)

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

(c)  \(y\!=\!-4f(x\!+\!2)\) involves both a vertical and a horizontal transformation. The \(y\)-coordinates are multiplied by \(-4\small,\) 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 \((1,\,4)\small.\)

(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 \((1,-2)\small.\)

(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 \((3,\,\frac{7}{2})\small.\)

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

$$ \begin{eqnarray} f\left(g(x)\right) &=& f\left(5-x\right) \\[6pt] &=& \frac{1}{\sqrt{5-x}} \end{eqnarray} $$

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

$$ \begin{gather} 5-x\leqslant 0\\[6pt] 5\leqslant x\\[6pt] x\geqslant 5\\[6pt] \end{gather} $$

(a)  Qualifications Scotland will accept two different methods of finding an inverse function.

Method 1:

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

$$ \begin{eqnarray} f\left(f^{-1}(x)\right) &=& x \\[6pt] \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{f^{-1}(x)}\,+\,8 &=& x \\[6pt] \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{f^{-1}(x)} &=& x-8 \\[6pt] f^{-1}(x) &=& (x-8)^3 \end{eqnarray} $$

Method 2:

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

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

$$ \begin{eqnarray} f(x) &=& \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{x}\,+\,8 \\[6pt] y &=& \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{x}\,+\,8 \\[6pt] y-8 &=& \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{x} \\[6pt] (y-8)^3 &=& x \\[6pt] x &=& (y-8)^3 \\[6pt] f^{-1}(y) &=& (y-8)^3 \\[6pt] f^{-1}(x) &=& (x-8)^3 \end{eqnarray} $$

(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 \(f\small.\)

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.

\(f(1)=\sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{1}+8=1+8=9\)
\(f(1000)=\sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{1000}+8=10+8=18\)

So the domain of \(f^{-1}\) is \(9\leqslant x\leqslant 18\small,\normalsize\ x\in\mathbb R\small.\)

Qualifications Scotland will accept two different methods of finding an inverse function.

Method 1:

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

$$ \begin{eqnarray} h\left(h^{-1}(x)\right) &=& x \\[6pt] 4+\small\frac{1}{3}\normalsize h^{-1}(x) &=& x \\[6pt] \small\frac{1}{3}\normalsize h^{-1}(x) &=& x-4 \\[6pt] h^{-1}(x) &=& 3(x-4) \end{eqnarray} $$

Method 2:

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

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

$$ \begin{eqnarray} h(x) &=& 4+\small\frac{1}{3}\normalsize x \\[6pt] y &=& 4+\small\frac{1}{3}\normalsize x \\[6pt] y-4 &=& \small\frac{1}{3}\normalsize x \\[6pt] 3(y-4) &=& x \\[6pt] x &=& 3(y-4) \\[6pt] h^{-1}(y) &=& 3(y-4) \\[6pt] h^{-1}(x) &=& 3(x-4) \end{eqnarray} $$

Qualifications Scotland will accept two different methods of finding an inverse function.

Method 1:

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

$$ \begin{eqnarray} f\left(f^{-1}(x)\right) &=& x \\[6pt] \small\frac{2}{f^{-1}(x)}\normalsize +3 &=& x \\[6pt] \small\frac{2}{f^{-1}(x)}\normalsize &=& x-3 \\[6pt] 2 &=& (x-3)f^{-1}(x) \\[6pt] f^{-1}(x) &=& \small\frac{2}{x-3}\normalsize \\[6pt] \end{eqnarray} $$

Method 2:

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

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

$$ \begin{eqnarray} f(x) &=& \small\frac{2}{x}\normalsize+3 \\[6pt] y &=& \small\frac{2}{x}\normalsize+3 \\[6pt] y-3 &=& \small\frac{2}{x}\normalsize \\[6pt] (y-3)x &=& 2 \\[6pt] x &=& \small\frac{2}{y-3}\normalsize \\[6pt] f^{-1}(y) &=& \small\frac{2}{y-3}\normalsize \\[6pt] f^{-1}(x) &=& \small\frac{2}{x-3}\normalsize \end{eqnarray} $$

Qualifications Scotland will accept two different methods of finding an inverse function.

Method 1:

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

$$ \begin{eqnarray} h\left(h^{-1}(x)\right) &=& x \\[6pt] 2\left(h^{-1}(x)\right)^3-7 &=& x \\[6pt] 2\left(h^{-1}(x)\right)^3 &=& x+7 \\[6pt] \left(h^{-1}(x)\right)^3 &=& \frac{x+7}{2} \\[6pt] h^{-1}(x) &=& \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{\frac{x\,+\,7}{2}} \\[6pt] \end{eqnarray} $$

Method 2:

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

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

$$ \begin{gather} h(x) = 2x^3-7 \\[6pt] y = 2x^3-7 \\[6pt] y+7 = 2x^3 \\[6pt] \frac{y+7}{2} = x^3 \\[6pt] \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{\frac{y\,+\,7}{2}} = x \\[6pt] x = \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{\frac{y\,+\,7}{2}} \\[6pt] h^{-1}(y) = \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{\frac{y\,+\,7}{2}} \\[6pt] h^{-1}(x) = \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{\frac{x\,+\,7}{2}} \\[6pt] \end{gather} $$

Two transformations have been applied: one horizontal and one vertical.

\(y=f(-x)\) represents a horizontal reflection, in the \(y\)-axis, so each \(x\)-coordinate changes sign.

The \(+3\) adds \(3\) to each \(y\)-coordinate, moving the graph up \(3\) units.

So the given points map as follows:

• The stationary point \((0,\,3)\) maps onto \((0,\,6)\small.\)
• The stationary point \((4,\,0)\) maps onto \((-4,\,3)\small.\)
• The root shown at \((-2,\,0)\) maps onto \((2,\,3)\small.\)

The resulting graph, after the horizontal reflection and vertical translation, is shown with the solid line below:

Qualifications Scotland will accept two different methods of finding an inverse function.

Method 1:

Use the definition of \(g^{-1}\) as the function for which \(g\left(g^{-1}(x)\right)=x,\) and rearrange:

$$ \begin{eqnarray} g\left(g^{-1}(x)\right) &=& x \\[6pt] \left(g^{-1}(x)-4\right)^3 &=& x \\[6pt] g^{-1}(x)-4 &=& \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{x} \\[6pt] g^{-1}(x) &=& \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{x}\,+\,4 \\[6pt] \end{eqnarray} $$

Method 2:

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

So we write \(g(x)\) as \(y\) and rearrange to make \(x\) the subject:

$$ \begin{gather} g(x) = (x-4)^3 \\[6pt] y = (x-4)^3 \\[6pt] \sqrt[\scriptstyle 3]{y} = x-4 \\[6pt] \sqrt[\scriptstyle 3]{y}+4 = x \\[6pt] x = \sqrt[\scriptstyle 3]{y}+4 \\[6pt] g^{-1}(y) = \sqrt[\scriptstyle 3]{y}+4 \\[6pt] g^{-1}(x) = \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{x}+4 \\[6pt] \end{gather} $$