Question

Hey guys, I'm trying to figure out how to "algebraically" prove there are no x-intercepts for graph of y = 1+sin 2x?... help please

Answer 1

it has x-intercepts

Answer 2

what ? how?... isn't it moved up 1 on the y axis with an amplitude of 1?

Answer 3

So instead of going up to 1 and down to -1,
it will go up to 2 and down to 0 (intercepting/touching the axis) yes?

Answer 4

You lose.

Answer 5

correct. but it's not necessarily intersecting right?

Answer 6

It `intercepts` the x-axis.
But yes you are correct it doesn't necessarily pass through which I assume is what you mean by `intersect`.

Answer 7

so if I set y= o and then solve for x i would get
0= 1+sin2x
-1 = sin 2x
sin = -1 @ pi/2 & 3pi/2 , so,
pi/2 = 2x
pi/4 =x

Answer 8

Only -1 at 3pi/2.

3pi/2 = 2x

Answer 9

oh,yea duh. so the only "intercept" is 3pi/4. correct?

Answer 10

No there is an infinite number of intercepts since sin2x is periodic.
I guess a better way to word it would be,
sin2x = -1 implies,\[\Large\bf\sf 2x\quad=\quad \frac{3\pi}{2}+2k \pi\]

Answer 11

no, there's actually infinitely many x-intercept

Answer 12

Where k is an integer, adding or subtracting full rotations gives us another intercept.

Answer 13

It's important to write the 2kpi BEFORE dividing by 2, because it affects the other solutions.

Answer 14

\[\Large\bf\sf x\quad=\quad \frac{3\pi}{4}+k \pi\]

Answer 15

When k=0, we get the expected solution 3pi/4.
When k=1, we get 7pi/4.
When k=2, we get 11pi/4.
And so on...

Answer 16

Confused by any of that? :o

Answer 17

I'm kind of getting confused, so I have to solve setting the 2x equal to the 3pi first? or thats only for the finding all other infinite solutions?

Answer 18

Setting 2x equal to 3pi/2 only finds the first solution.
Setting 2x equal to 3pi/2 + 2pi is the second solution.
Understand how those two solutions would fall on the same spot on the unit circle but the second one has rotated around a full time?