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2015 II Q4 ii,iii)

I'm not exactly sure how to get to y=g(x) or y=h(x)?

I mean, I can sketch y in both cases, and I get that it's the arc tan function, but I'm just not 100% sure how to get there

I kind of get the gist but I'd like a 100% solid understanding if you know what I mean?

Thanks in advance!

Not really sure what you're asking here. But in part (ii) for instance (part (iii) is pretty much the same process) you simply sketch $y = \frac{x}{1+x^2}$ and then see how that looks like. Then you sort of treat that curve as your $x$-axis in plotting $g(x) = \arctan y$.

Since you know that $g(0) = \pi$ then that gives you one point on the graph. You also know that $y$ has extrema at $x=\pm 1$ and since $\arctan$ is increasing, then $\arctan y$ will have extrema at $x=\pm 1$ as well.

That is, $y$ is maximised at $x=1$ with value $1/2$ so $g(x)$ will be maximised at $x=1$ with value $\arctan (1/2) = \pi + \pi/4$. Similarly minimised at $x=-1$ with value $\pi - \pi/4$.

That gives you the general shape of the graph you want to sketch and the final clinch is just basic observations of the asymptotes. This is how you sketch graphs in general, simply look at the extrema/asymptotes/intercepts, etc...

$\arctan\left (\frac{1}{2} \right ) \neq \frac{\pi}{4}$

(Actually, I haven't said that $\arctan \frac{1}{2} = \frac{\pi}{4}$ - I made a different mistake)

But thanks, I've caught the error. Should be $g$ max/min at $\pi \pm \arctan \frac{1}{2}$ respectively.

you originally said the answer was $\pi \pm \frac{\pi}{4}$ but now you're saying it's $\pi \pm \arctan ( \frac{1}{2} )$. if you didn't mistakenly use $\arctan ( \frac{1}{2} ) = \frac{\pi}{4} $, I'm not sure I understand your original solution

thanks in advance

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