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

[seniorhs9]

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

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.

 

[SammyS]

...
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.

 

[seniorhs9]

Hi SammyS. Thanks for your answer.

But I don't understand your answer either

"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]

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.

 

[seniorhs9]

Hi SammyS. Thanks.