How to find x-intercepts of the graph of $f$

Question

Given the following function $$f(x) = \frac {\sin (4x)}{2x}$$

How to find $x$-intercepts of the graph of $f$ on the interval $0 \le x \le \pi$?

score 1 · Answer 1 · answered Sep 08 '13 at 05:10

1

For x-intercepts, $f(x)=0$

Observe that at $\displaystyle x=0,f(x)=\frac00$ i.e., undefined

So, we need $\sin4x=0,x\ne0\implies 4x=n\pi$ where $n$ is any integer$\ne0$

$\implies x=\frac{n\pi}4$

Now for the given interval, $0<\frac{n\pi}4\le \pi\iff 0<n\le 4$

answered Sep 08 '13 at 05:10

lab bhattacharjee

I'm still confused, sorry. Can you elaborate further? – Grey Sep 08 '13 at 05:33
@Grey, sure. $x$ intercept means when $f(x)$ intersects with $x$ axis whose eqaution is $y=0$ – lab bhattacharjee Sep 08 '13 at 05:35
Is it x=(πn/2) and x=1/4(2πn+π) ? – Grey Sep 08 '13 at 05:49
@Grey, not sure how you have found this, but observe that this is same as $x=n\frac{\pi}4$ – lab bhattacharjee Sep 08 '13 at 05:50
I'm not sure either. Anyway, thanks for the help. – Grey Sep 08 '13 at 05:54
@Grey, my pleasure. Hope you understood my method – lab bhattacharjee Sep 08 '13 at 05:55

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