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 \(x\).

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

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

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

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

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

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

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

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

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

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

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

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:

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

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

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

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

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

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

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

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