Math Help - State four points where y = cosx = y = (1/2)^2

1. State four points where y = cosx = y = (1/2)^2

y = cosx, y = (1/2)^x

- State four points where cosx = (1/2)^2
- Also, why are there an infinite number of these points?

I've gotten the graphs but except for x = 0, I don't know which other points to choose. The graph can be seen below:

I need to get this in by tomorrow so if any of you wonderfolks can help me out with this, that'd be great. Thanks ofcourse!

On a sidenote: Would it be right to say that sinx ≤ x when x > 0?

2. Re: State four points where y = cosx = y = (1/2)^2

Well (1/2)^2 = 1/4, so you should just be ruling a horizontal line at y = 1/4 and show that it crosses the graph y = cos(x) infinitely many times.

3. Re: State four points where y = cosx = y = (1/2)^2

That was a typo- he meant $cos(x)= (1/2)^x$. Bilal7, since $(1/2)^x$ is always positive it will cross cos(x) an infinite number of times. You will have to use a numerical method to approximate solutions. You could, for example, use "Newton-Raphson": to solve $f(x)= cos(x)- (1/2)^x= 0$ take the derivative: $f'(x)= -sin(x)- ln(2)(1/2)^x$ and form the sequence $x_{n+1}= x_n- \frac{f(x)}{f'(x)}$, choosing a starting $x_0$ from your graph that is reasonably close to a root.

4. Re: State four points where y = cosx = y = (1/2)^2

Originally Posted by Bilal7
y = cosx, y = (1/2)^x

- State four points where cosx = (1/2)^2
- Also, why are there an infinite number of these points?

I've gotten the graphs but except for x = 0, I don't know which other points to choose. The graph can be seen below:

I need to get this in by tomorrow so if any of you wonderfolks can help me out with this, that'd be great. Thanks ofcourse!

On a sidenote: Would it be right to say that sinx ≤ x when x > 0?
Let's start with the side note. $0 < x \implies sin(x) < x\ and\ 0 \le x \implies sin(x) \le x.$ Proof follows.

$Let\ z = x - sin(x).$

$sin(x) \le 1\ \implies - 1 \le - sin(x) \implies 0 < x - sin(x)\ if\ 1 < x \implies sin(x) < x\ if\ 1 < x.$

$sin(0) = 0 \implies sin(x) = x\ for\ x = 0.$

$Assume\ 0 < x \le 1.$

$So\ cos(x) < 1 \implies - cos(x) > - 1 \implies 1 - cos(x) > 0.$

$z = x - sin(x) \implies \dfrac{dz}{dx} = 1 - cos(x) > 0 \implies z\ is\ increasing \implies z > 0 \implies x - sin(x) > 0 \implies sin(x) < x.$

As for your main question, I have no idea what it is. What you say in your title and what you say in your text are different. And are you being asked for exact answers or approximate answers? Please state the problem completely and exactly.

5. Re: State four points where y = cosx = y = (1/2)^2

Yes, there was a typo in the thread title and the second function is infact y = (1/2)^x, and the question doesn't state if they should be exact or approximate answers so I guess either would be fine.

@Halls of Ivy- We haven't covered that "Newton-Raphson" method so I went back and used a graphing tool and here are the results:

Point 1: (0, 1)
Point 2: (1.08, 0.47)
Point 3: (4.75, 0.04)
Point 4: (7.85, 0)

Look good? This is how they did it in the textbook too. Also, there are an infinite number of points because y=(1/2)^x is always positive, right?

6. Re: State four points where y = cosx = y = (1/2)^2

Originally Posted by Bilal7
Yes, there was a typo in the thread title and the second function is infact y = (1/2)^x, and the question doesn't state if they should be exact or approximate answers so I guess either would be fine.
@Halls of Ivy- We haven't covered that "Newton-Raphson" method so I went back and used a graphing tool and here are the results:
Point 1: (0, 1)
Point 2: (1.08, 0.47)
Point 3: (4.75, 0.04)
Point 4: (7.85, 0)
Here is a better plot.

BUT, I really am curious about the point of this question?
What concept and/or principle is the author trying to make?
Surely it is nothing to do with the period of $\cos(x)~???$

7. Re: State four points where y = cosx = y = (1/2)^2

Originally Posted by Plato
Here is a better plot.

BUT, I really am curious about the point of this question?
What concept and/or principle is the author trying to make?
Surely it is nothing to do with the period of $\cos(x)~???$
To be honest, I fail to see the point behind several math problems. No one is going to be using this trignometry stuff if they're not going into architecture. :P

Also, your plot uses 2^x while the question says (1/2)^x.

8. Re: State four points where y = cosx = y = (1/2)^2

Originally Posted by Bilal7
No one is going to be using this trigonometry stuff if they're not going into architecture.
Also, your plot uses 2^x while the question says (1/2)^x
.
None of that is true.

Your real problem is your failure to have a basic grasp of the fundamentals.
If you did knew the basics, you would have seen that ${\left( {\dfrac{1}{2}} \right)^x} = {2^{ - x}}~.$