# Meaning of y = c if f(x) = cg(x)

## Homework Statement

Hi. I post part A too but I'm having trouble only with Part B.

Question.[PLAIN]http://img100.imageshack.us/img100/8297/20104ivwhytangenttoseco.png [Broken]

## Homework Equations

I know....

$y = sin x$ and $y = cx$ intersects when ...

$sin x = cx$ so $\frac{sin x}{x} = c$

## The Attempt at a Solution

I just don't understand why $\frac{sin x}{x} = c$
means that $y = c$ has to be tangent to
$y = \frac{sin x}{x}$ at the second hump?

I'm not seeing the connection...

Thank you.

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SammyS
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Homework Helper
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...
I know....

$y = \sin x$ and $y = cx$ intersects when ...

$\sin x = cx$ so $\displaystyle \frac{\sin x}{x} = c$

## The Attempt at a Solution

I just don't understand why $\displaystyle \frac{\sin x}{x} = c$
means that $y = c$ has to be tangent to
$\displaystyle y = \frac{\sin x}{x}$ at the second hump?

I'm not seeing the connection...

Thank you.
Look at the second graph. If y = c x is not tangent to the graph of y = sin(x) "near the second hump", then the graph of y = c x will intersect the graph of y = sin(x) in either 7 places, or in 3 places.

"If y = cx is not tangent to the graph of y = sin(x) "near the second hump", then the graph of y = cx will intersect the graph of y = sin(x) in either 7 places, or in 3 places "
Generally, I have trouble relating intersection of $y = sin x$ and $y = cx$ to graph of $y = \frac{sin x}{x}$ and $y = c$. But I understand $sin x = cx$ means $\frac{sin x}{x} = c$.

SammyS
Staff Emeritus
Homework Helper
Gold Member
To solve y = c x , and y = sin(x) simultaneously, you have:

c x = sin(x) .

That's equivalent to $\displaystyle c=\frac{\sin(x)}{x}\,.$

One way to solve this equation is graphically, i.e., where does the graph of y = c intercept the graph of $\displaystyle y=\frac{\sin(x)}{x}\,?$ I think the question to be asking is, "Should these two graphs be tangent at the point of intersection?"

If y = c is tangent to $\displaystyle y=\frac{\sin(x)}{x}\,$ near $\displaystyle x=\frac{5\pi}{2}\,,$ then it must be true that on either side of the point of tangency, $\displaystyle c>\frac{\sin(x)}{x}\,.$ It can be shown that this inequality holds.

Therefore, the slope of $\displaystyle y=\frac{\sin(x)}{x}\,$ is zero at the same value of x at which the the graph of y = c x is tangent to the graph of $\displaystyle y=\sin(x)\,$ near the second hump. Use this value of x to solve for c.

Hi SammyS. Thanks.