Hi again,

Can someone please help me with q3 and q6 b.

Thanks,

beep boop

Hi beep boop

**Question 6**\(\mathbf{b.}\quad\) Note make sure you look back at what you started with. The the maximal domain of \(f\left(x\right)\) is \(x\in \left(0,\infty \right)\) (this is because \(\log _{10}\left(x\right)\) is only defined for \(x>0\)).

Now assuming you were able to yield to the result in part a, solve \(x=f\left(y\right)\) with consideration of the above domain.

\(\frac{y^2}{10y+10}=x\:\Longrightarrow \:y^2=x\left(10y+10\right)\Longrightarrow y^2-10xy-10=0\)

From here you can apply quadratic formula where \(a=1,\:b=-10x,\:c=-10\)

This will yield to \(y=\frac{10x+\sqrt{100x^2+40x}}{2}\:\text{or}\:\frac{10x-\sqrt{100x^2+40x}}{2}\).

However, you need to consider which solution satisifies conditons of \(f\left(x\right)\) or in other words, which solution has a range of \(\left(0,\infty \right)\). (since \(\text{dom}\:f=\text{ran}\:f^{-1}\). Testing out \(x=1\) for both of the solutions show that the first solution is the inverse function.)

Therefore, \(f^{-1}:\:\left(0,\infty \right)\rightarrow \mathbb{R},\:f^{-1}\left(x\right)=\frac{10x+\sqrt{100x^2+40x}}{2}\)

We got the domain of \(f^{-1}\) to be \(\left(0,\infty \right)\) because that is the range of the original function

**Question 3**\(\mathbf{a.}\quad\) For the purpose of my explanation suppose \(g\left(x\right)=f\left(x-5\right)\), then \(g\left(x+5\right)=f\left(\left(x+5\right)-5\right)=f\left(x\right)\). In other words just substitute \(x+5\) into \(f\left(x-5\right)\) to obtain original function.

\(\mathbf{b.}\quad\) Like above, lets suppose \(g\left(x\right)=2f\left(-2\left(x-\frac{1}{2}\right)\right)+5\). Therefore, \(\frac{1}{2}g\left(x\right)-\frac{5}{2}=f\left(-2\left(x-\frac{1}{2}\right)\right)\) and by solving \(-2\left(x-\frac{1}{2}\right)=x'\) you can find the horizontal transformations needed to undo it. (that is dilation of factor \(\frac{1}{2}\) from \(y\) axis, reflection in \(y\) axis and translation of \(1\) unit right.) Therefore, \(\frac{1}{2}g\left(-\frac{x-1}{2}\right)-\frac{5}{2}=f\left(x\right)\). Now substitue that into the equation. (I'm assuming this quesiton is tech-active so pop it in your CAS!)

Hope this helps