National 5 Maths
Functions

Here, we are being asked to find the output if the input is \(5.\)

So we substitute \(5\) in place of the general input variable \(x\).

$$ \begin{eqnarray} f(5) &=&\ 2\times 5 + 3\\[6pt] &=&\ 10 + 3\\[6pt] &=&\ 13 \end{eqnarray} $$

Video solution by Clelland Maths 

We have to find the output if the input is \(-4.\)

So we substitute \(-4\) in place of the general input variable \(x\).

$$ \begin{eqnarray} f(-4) &=&\ 7-3(-4)\\[6pt] &=&\ 7+12\\[6pt] &=&\ 19 \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} f(6)\ &=&\ 4(6^2)-5(6)+ 2\\[6pt] &=&\ 4(36) - 30 + 2\\[6pt] &=&\ 144 - 30 + 2\\[6pt] &=&\ 116 \end{eqnarray} $$

Video solution by Clelland Maths 

It would be a waste of time to expand the brackets. We can substitute directly:

$$ \begin{eqnarray} f(7)\ &=&\ (7+3)(7-5) \\[6pt] &=&\ 10 \times 2\\[6pt] &=&\ 20 \end{eqnarray} $$

Video solution by Clelland Maths 

We are being told that the output is 29. We have to find the matching input.

$$ \begin{eqnarray} f(a)\ &=&\ 29\\[6pt] 5a - 1\ &=&\ 29\\[6pt] 5a\ &=&\ 30\\[6pt] a\ &=&\ 6 \end{eqnarray} $$

Video solution by Clelland Maths 

We are being told that the output is 8. We have to find the corresponding input.

$$ \begin{eqnarray} f(t)\ &=&\ 8\\[6pt] \sqrt{3t+4}\ &=&\ 8\\[6pt] 3t+4\ &=&\ 8^2\\[6pt] 3t+4\ &=&\ 64\\[6pt] 3t\ &=&\ 60\\[6pt] t\ &=&\ 20 \end{eqnarray} $$

Video solution by Clelland Maths 

We are being told that the output is 9. We have to find the corresponding input.

What's tricky here is that the input variable is the denominator. So the first step of our working is to multiply through by \(a\).

$$ \begin{eqnarray} f(a)\ &=&\ 9\\[6pt] 1+\small\frac{4}{a}\normalsize\ &=&\ 9\\[6pt] a+4\ &=&\ 9a\\[6pt] 4\ &=&\ 9a-a\\[6pt] 8a\ &=&\ 4\\[6pt] a\ &=&\ \small \frac{4}{8}\normalsize \\[6pt] a\ &=&\ \small\frac{1}{2}\normalsize \end{eqnarray} $$

There are other equally sensible ways of answering this question. Perhaps you prefer this working:

$$ \begin{eqnarray} f(a)\ &=&\ 9\\[6pt] 1+\small\frac{4}{a}\normalsize\ &=&\ 9\\[6pt] \small\frac{4}{a}\normalsize\ &=&\ 9-1 = 8\\[6pt] 4\ &=&\ 8a\\[6pt] 8a\ &=&\ 4\\[6pt] a\ &=&\ \small \frac{4}{8}= \small\frac{1}{2}\normalsize \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} f(a)\ &=&\ 23\\[6pt] 3a+2\ &=&\ 23\\[6pt] 3a\ &=&\ 23-2\\[6pt] 3a\ &=&\ 21\\[6pt] a\ &=&\ 7 \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} f(-5) &=&\ (-5)^2+3(-5)\\[6pt] &=&\ 25-15\\[6pt] &=&\ 10 \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} f(a)\ &=&\ 73\\[6pt] 5+4a\ &=&\ 73\\[6pt] 4a\ &=&\ 73-5\\[6pt] 4a\ &=&\ 68\\[6pt] a\ &=&\ \small\frac{68}{4}\\[6pt] a\ &=&\ 17 \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} f(-2) &=&\ 5(-2)^3\\[6pt] &=&\ 5(-8)\\[6pt] &=&\ -\!40 \end{eqnarray} $$

Video solution by Clelland Maths 

This non-calculator question combined functions with trig graphs. You need to know that \(sin\,270^\circ = -1.\)

$$ \begin{eqnarray} f(90) &=&\ 4\,sin\,\left(3\times 90\right)^\circ\\[6pt] &=&\ 4\,sin\,270^\circ\\[6pt] &=&\ 4\times -1\\[6pt] &=&\ -\!4 \end{eqnarray} $$

Video solution by Clelland Maths 

$$ \begin{eqnarray} f(-3) &=&\ \left(-3\right)^3-2\\[6pt] &=&\ -\!27-2\\[6pt] &=&\ -\!29 \end{eqnarray} $$

Video solution by Clelland Maths 

Multiplying out the brackets would be a waste of time and lead to more complicated working. It is much better to substitute directly into the expression, as it is.

$$ \begin{eqnarray} f(7) &=&\ \left(7+3\right)^2\\[6pt] &=&\ 10^2\\[6pt] &=&\ 100 \end{eqnarray} $$

Video solution by Clelland Maths 

(a)  We just substitute \(x=6\) into the function. $$ \begin{eqnarray} f(6) &=&\ 3\times 6+7\\[6pt] &=&\ 18+7\\[6pt] &=&\ 25 \end{eqnarray} $$ (b)  We work backwards from knowing the value of \(f(p)\small.\)
$$ \begin{eqnarray} f(p)\ &=&\ 19\\[6pt] 3p+7\ &=&\ 19\\[6pt] 3p\ &=&\ 19-7\\[6pt] 3p\ &=&\ 12\\[6pt] p\ &=&\ \small\frac{12}{3}\\[6pt] p\ &=&\ 4 \end{eqnarray} $$