<p>How many times does the graph y=0.1x intersect the graph of y=sin(2x)?</p>
<p>errgh, it seems so easy but i can't seem to solve it!</p>
<p>I think the answer has to be 1
The 2 graphs intersect only at the origin</p>
<p>... graph them on your calc, the upper part of the sin graph will intersect with the line in the -1>y>1 range for the line. count the total amount of intersections on your graphing calc and its 11 I think. probably solvable by hand but this way seems so much faster in this case.</p>
<p>yeah lynda is right , I had my calc on deg instead of rad...
11 is the correct answer</p>
<p>How would you answer
How many times does the graph y=0.01x intersect the graph of y=sinx?</p>
<p>Gcf101:
...or to put your question another way: 'How would you answer this without a graphing calculator?'</p>
<p>HINT:
a) What's the value of x when y=1? 100, right?
b) What's 100/(2 pi) ?</p>
<p>Thanks for hinting!</p>
<p>Is it 127 intersections??</p>
<p>100/pi=31,83 periods
2 intersections per 1 period
so 2*31= 62
but because of 0,83 another intersection, so 62+1 intersections on the right side of the Y.
So on both side there are 63+63=126
And another one at the origin, so in total 127</p>
<p>Actually, the origin is counted twice (once for [0,2 pi] and again for [-2 pi,0]. Also, you need 100/2pi (=15.91) , not 100/pi .</p>
<p>A harder question to answer is how many intersections there are (if any) in the interval <a href="and%20in%20the%20symmetric%20interval%20on%20the%20other%20side%20of%20the%20y-axis"> 30 pi, 31.83 pi</a>. sin( 30 pi)=0, and sin(30.5 pi) = 1 ; so we have extra intersections in [30 pi, 30.5pi] and in [-30.5 pi, -30pi]. sin(x) drops to 0 as we move from x=30.5pi to x=31pi, so there's another intersection here (and in [-31 pi, -30.5 pi]).</p>
<p>I think the correct answer is (2)(2)(15) -1 + (2)(2) = 63.</p>
<p>Why do you divide By 2pi, the period of sin(2x) is actually pi...
And for every period there is 2 intersections</p>
<p>sin(x) goes from 0 -> 1 -> 0 -> -1 -> 0, as x goes 0 -> pi/2 -> pi -> 3pi/2 -> 2pi. There is (usually) 1 intersection in the interval [ 2n * pi, 2n * pi + pi/2], and another in the interval [ 2n * pi + pi/2, (2n+1) * pi], for n=0.1,2,.... . However, in the interval [(2n+1) pi, (2n+2) pi], sin(x) is <= 0, and will have no intersection with y=0.01 x . </p>
<p>A similar argument holds for n=-1,-2,-3,.. .</p>
<br>
<p>How many times does the graph y=0.1x intersect the graph of y=sin(2x)?</p>
<br>
<p>sin(2x) <= 1, for all x
therefore
for all 0.1x >1, or x>10
then 0.1x and sin(2x) do not intersect (not equal)
And
sin(2x) at x=10 is sin(20)
And
3<em>(2pi) < 20 < 4</em>(2pi)
therefore
y=0.1x intersects twice with 3 positive half-cycles of sin(2x) for x=>0
y=0.1x intersects twice with 3 negative half-cycles of sin(2x) for x<=0
however, two intersections overlap at 0
0.1(0)=0=sin(2<em>0)
therefore
instead of (2</em>3) + (2*3) = 12, there are only 11 intersections</p>
<p>(1/16)^x = log (base 1/16) (x)</p>
<p>How many solutions does this equation have?</p>
<p>Gcf101:
You like to bust our collective chops, don't you?</p>
<p>As x goes 0 -> infinity, y<em>1 = (1/16)^x should go 1 -> 0 . Since LHS= RHS in your equation, (1/16)^LHS = (1/16)^RHS or (1/16)^y</em>1 = x.</p>
<p>As x goes 0 -> infinity, y<em>1 goes 1->0 and (1/16)^y</em>1 should go (1/16) -> 1 in a smooth curve. I suspect this should intersect y<em>2 = x at one point only, somewhere in the range (1/16) <= x <= 1. Unless y</em>1 'wiggles' sinusoidally, which I doubt.</p>
<p>OK, what's the catch?</p>
<p>chochocho: where did you find your question?</p>
<p>optimizerdad:
this question (post #13) blew my mind many years ago, and I was happy to share it when somewhat similar question (post #1) came up.</p>
<p>I'll be brief (SAT related business takes most of my time now).
Your approach is very interesting, but a little overcomplicated.
Everything you deduced is correct except for the last part.
The fact that
y_1 = (1/16)^x does not "wiggle"<br>
from concave upward to concave downward,
does not mean that
y=(1/16)^(1/16)^x can't wiggle.
Is not that amazing?!</p>
<p>This link might be useful:
<a href="http://www.math.hmc.edu/calculus/tutorials/secondderiv/%5B/url%5D">http://www.math.hmc.edu/calculus/tutorials/secondderiv/</a></p>
<p>Using TI-89, you can find that for
y=(1/16)^(1/16)^x
y''=0 at some point at x<em>0, and y'' changes sign at x</em>0, which is then a point of inflection, or "wiggling" (like x=0 for y=sin x).</p>
<p>So it's not impossible that
y=(1/16)^(1/16)^x
intersects y=x at more then one point.
Sorry!</p>
<p>Here's the question again:
(1/16)^x = log (base 1/16) (x)
How many solutions does this equation have?</p>
<p>Stick to the graphs
y=(1/16)^x and
y= log (base 1/16) (x);
also do some guess and check.
TI-89 "solve" feature might help to the extent.
Correct solving by graphing and zooming is probably impossible.</p>
<p>Just a reminder to those not solid in logs:
log(base m) (x) = n means
m^n = x (this is actually a definition of the log function).
Put into words:
log(base m) (x) is a number such that if you raise a base (nice rhyme!) to "it" you'll get "x".
In other words:
m^(log(base m) (x)) = x.</p>
<p>Examples:
log (base 2) (16) = 4 because 2^4=16.
log (base 1/125) (1/5) = 1/3 since (1/125)^(1/3) = 1/5.</p>
<p>(1/125)^(log(base125) (1/5)) = 1/5.</p>