How to draw a graph of sin(x)/x

Or any other product function for that matter.
I have no idea how I would go about this (without a graphical calculator/google)
I have an interview in a week or so and I feel like I should know how to graph something like this.
How would I do this?

Reply 1

NMcG

Original post by Someboady

Start by concluding x can never = 0 then just plot some points that you know from a normal sin graph, 30,45,60,90 etc?

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Reply 2

BasicMistake

Follow this general rule:

1. Where does the curve intercept the axes?
2. Does the curve have asymptotes and if so, where?
3. Find the turning points by differentiating the equation

Reply 3

Someboady

Original post by BasicMistake

Follow this general rule:

1. Where does the curve intercept the axes?
2. Does the curve have asymptotes and if so, where?
3. Find the turning points by differentiating the equation

I'm not sure where it intercepts the y-axis? If I was to make x = 0, then there would be division by zero so would that mean there is an asymptote at x=0?
However when I look at the graph it goes through (0,1) and I'm not sure why...
I think it has too many turning points to differentiate and figure out?

Reply 4

djmans

Original post by Someboady

did you try searching for sin(x)/x on google, it gives you a graph not sure if it is corrrect

Reply 5

RDKGames

Study Forum Helper

Original post by Someboady

It does not go through (0,1) - this graph has no y-intercept as it is discontinuous at x=0.

You can indeed find the limit as

x\rightarrow 0

by applying L'Hopital's rule or using MacLaurin's expansion on

\sin(x)

in order to find what value the graphs approaches as x goes to 0 - realistically your problem is

\displaystyle \lim_{x\rightarrow 0} \frac{\sin(x)}{x}

You can tell this is a discontinuous graph because your denominator can be equal to 0, in which case you need to find the limits to which the graph approaches as x tends to its asymptotic values before sketching it. This is easier with rational functions as the discontinuity can be cancelled in some cases which makes finding the limit easy. You can indeed find this to be true if you were to use the expansion and divide through by x.

As for sketching the rest of it, just find the stationary points and go from there - a sufficient domain to sketch this on would be

-2\pi \leq x \leq 2\pi

(edited 7 years ago)

Reply 6

Someboady

Original post by RDKGames

It does not go through (0,1) - this graphs has no y-intercept as it is discontinuous.

You can indeed find the limit as

x\rightarrow 0

by applying L'Hopital's rule or using MacLaurin's expansion on

\sin(x)

- initially your problem is

\displaystyle \lim_{x\rightarrow \infty} \frac{\sin(x)}{x}

You can tell this is a discontinous graph because your denominator can be equal to 0, in which case you need to find the limits to which the graph approaches as x tends to its asymptotic values before sketching it.

ahhhhh! See I used my graphical calculator and it drew a graph that passes through (0,1) and I figured it was asymptotic at x=0 so I was slightly confused...
I was told there is a way of drawing the graph by expressing it as a product of two functions i.e. sinx and 1/x and then I'm not sure what comes next. How would you draw the graph using this?

Reply 7

Plagioclase

Original post by Someboady

That method works well for some functions, you draw the two graphs and then use your intuition to work out the shape of the product, e.g. you can very easily work out when the resulting function will be positive or negative, where roots are, etc.

This approach will allow you to work out how most of the sin(x)/x graph looks like, however it breaks down at x=0 where you need to do a bit more thinking and find the limit as x approaches zero.