Hey
It's important to show us what you've tried in future questions - it helps further your understanding more than us just giving you the answers. That being said, here are a few hints:
Q1
=f(-x) \implies \frac{p(-x)+q}{(-x)+r} - \frac{px+q}{x+r} = 0<br />\\ \frac{(q-xp)(x+r) - (px+q)(r-x)}{r^2-x^2} = 0<br />\\ 2qx - 2pxr= 0<br />\\ q = pr<br />\\ \text{Subbing back into our original definition of } f(x), f(x) = \frac{px+pr}{x+r} = p<br />\\ \text{as stipulated.}<br />)
Notice that this tells you that if the function is even, the function is constant!
The logic for Q2 follows much the same line as Q1 - you assume that \(-f(x) = f(-x)\) and then work through some similar algebra as the above to find \(f(x)\) in terms of q. If you really can't work through this part, let us know
What Q3 is essentially asking is for the inverse function of \(f(x) = \frac{3x+8}{x-3}\) ie. find a function such that \(x = \frac{3y+8}{y-3}\), but rearranged so \(y\) is a function of \(x\). It also asks for solution(s) to \(\frac{3x+8}{x-3} = x\) which is going to be a quadratic that you have to solve - again if you really can't work through this, let us know
Hope this helps
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