Sketching functions

First of all try to figure out if the function is odd, even or neither.
To do that, check if f(-x) = f(x). If that's the case, then the positive part of the function is a mirror image of the negative part. If f(x) = - f(-x) then the positive part is identical to the negative part but reflected over the x axis. If none of the above is true, then the function behaves differently on either ends of the y-axis.

Secondly, you notice that sinx is a sinusoidal function, that varies between positive and negative. Therefore, as x increases above zero, you f(x) will take positive and negative values sinusoidally. However, you also notice that sin(x) is actually multiplied by e^x, which means that the peaks of the sine curve increase as x increases, so you get a wave that grows on the positive end of the graph.

Thirdly, you check the negative part, is it going to be similar? Well, returning to point 1, if you plug in f(-x), you get e^-x * sin(-x) = -e^-x*sin(x)
As you can see, the graph is neither even nor odd. So you expect it to behave differently on the negative end.
Why's that, you should ask. Well, if you look at the graph of e^x, as x becomes more negative, e^x falls in value, but always remains positive. So you fundamentally you still have a sine curve, but the peaks keep falling in height because you're multiplying by smaller numbers ( e^x) as you go the negative end of the x-axis.

Fourthly, you can try to figure out where the graph meets the x-axis, that's really simple, it's basically the values of x where sin(x) is zero.

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As for the rest of your questions, try what we just did. However, notice that if f(x) = x cosx, then f(-x) = (-x) cos(-x) = - x cosx = - f(x) so the function is odd, i.e. the negative bit is just a mirror image of the positive bit over the y- axis.
See image:

So many people have told me that at Oxbridge interviews, a graph question is more than likely to come up. I was doing some practice on rational functions and I am fine with that but then I came across

When it comes to sketching functions, I already differentiate them to find any stationary points. Depending on the function, I can also find any critical points and test on either side of them to see from which direction (+/-) the function is coming from.

$\frac{d}{dx}e^xsin(x)=e^xcos(x)+e^xsin(x)=e^x(sinx+cosx)=0$
Therefore $sin(x)=-cos(x)$ which gives $tan(x)=-1$ and as you can see there will be an infinite amount of stationary points but really I suggest you take the ones between $-2\pi<x<2\pi$ which can be our critical values.
This also tells us that this is a wave type looking graphs so we have a rough idea of what it would look like.

Next see at what points the graph is 0, and this is obviously for $x=k\pi$ due to sine. You should also set yourself a domain on which you initially graph, such as $\pm2\pi$ as it's with trig. Test on either side of to get an idea from which direction the graph should come from.

At this point you should have enough information as to sketch it within a reasonable domain around x=0.

That's really unnecessary in most situations. Sometimes you have functions that are a pain to differentiate, but can very easily be understood by just playing around with the function. Testing the function at zero, infinities, obvious critical points, and so on.

Each to their own I suppose. I'm just used to differentiating if I believe I can and it would lead to anything meaningful.

Well actually come to think about it in a lot of situations you have to differentiate to find peaks and what not, but try to avoid it as much as I can

Yep, though in some cases, as you said, it's best to avoid it. Like in OP's post with $y=xsinx$. The differential leads to a transcendental equation if you attempt to solve for 0 which can't be done algebraically.

@OP, some unsolicited advice: graph sketching does come up in Oxbridge interviews (although mine didn't have any) but it's more in the general context of having to problem solve your way through them, how to intuit or explore different characteristics of the function, it's limiting value, parity, extrema, etc... - it's a bit pointless asking for just guidelines on how to sketch functions, they're not interested in you learning a set of pre-determined rules and applying that; they're interested in how you approach problems. A realistic graph-sketching problem that would trip you right up is to sketch the set of points $(x, y)$ which satisfy $x^y = y^x$, for positive $x,y$; rather different from what you're used to, I'd wager.

So really, I'd focus on your problem solving skills in general, if I were you, otherwise you don't stand much of a chance.

While I agree a 100%, I still do think learning some base methods is quite helpful, especially if he's not applying to do maths.

My cousin's doing NatSci and in his interview the interviewer asked him to sketch f(x) = sinx/x as well as f(x) = e^sin(x), which are both pretty standard that you could do with some base methods. But I do agree, learning by rote is a bad way to prepare for an interview at Oxbridge, if not accompanied by some solid problem solving skills learning.

Sounds like you are grovelling to me.

I am merely disagreeing with the conjecture that learning some basic formulae to solving problems isn't necessarily helpful in an Oxbridge interview. I think that rote learning is definitely not a good way to go about learning mathematics, but learning some standard methods is usually helpful especially if the candidate is not applying for maths directly.
I did not know that disagreeing with someone but not entirely has become a sycophantic thing to do these days.

sycophantic?
Sorry no dictionary

Same meaning as grovelling if I am not mistaken

So many people have told me that at Oxbridge interviews, a graph question is more than likely to come up. I was doing some practice on rational functions and I am fine with that but then I came across

While I agree a 100%, I still do think learning some base methods is quite helpful, especially if he's not applying to do maths.

Yep, was assuming he was applying for maths; checked his profile out and think he's applying for NatSci instead, oh well - you can disregard my advice OP, it doesn't apply anymore. :-P