Thread: How to show graphically that sinx=mx has no non-zero solutions...

The question is:
show graphically that sinx=mx has no non-zero solutions for -pie<x<pie when m<0.

Solution:
I drew the two graphs just by estimates,and they dint intercept.As the m is always going to be steeper,but I wanted to know if there is a concrete way to prove this.You know like doing a table or something.

I know it is a very nooby question,and I apologize for the inconvenience.
But honestly,with my Uni exams in a month and me missing most of the classes,this forum is my only source of getting ANY problems sorted out and hence I came here.

Thanks in advance for the help.

You are given m<0.
Consider two regions:
:Here sin(x) is positive and mx is negative,hence they don't intersect.
:Here sin(x) is negative and mx is positive,hence they don't intersect.
Only intersection point is 0.

Originally Posted by ignite

You are given m<0.
Consider two regions:
:Here sin(x) is positive and mx is negative,hence they don't intersect.
:Here sin(x) is negative and mx is positive,hence they don't intersect.
Only intersection point is 0.

thanks,could you solve one more problem for me?
i don't know if I should put up another thread for this,so posting it here anyways.
Here is the question:
by considering the case x>=0,-1<x<0 and x<=-1 prove that 1+x+x^2+x^3+x^4 is always positive.

Attempt at solution:
I did,f(0),F(1),F(-1),F(-0.5) and F(-2) and it all came positive,and so I said it is proven,am I right or is there something more I should do?

Also they had asked me before to show that 1+x+x^2 is always positive,I did that by completing square method.

The part c asks me to generalize the answers.What Should I write?

Edit:
I wrote:
All Polynomials with all coefficients equal to 1 and the constant equal to 1,will always be positive,regardless of their degree.

Originally Posted by syedmaheen

thanks,could you solve one more problem for me?
i don't know if I should put up another thread for this,so posting it here anyways.
Here is the question:
by considering the case x>=0,-1<x<0 and x<=-1 prove that 1+x+x^2+x^3+x^4 is always positive.

Attempt at solution:
I did,f(0),F(1),F(-1),F(-0.5) and F(-2) and it all came positive,and so I said it is proven,am I right or is there something more I should do?

No!

What you did was just to check functions value at particular points.You have to prove in general.
Case I: You can easily see that as all
Case II: In this case and are positive while and are negative.But and
Case III: Here , ,

Your generalisation is almost correct,except for the degree part.Think about it again.

Originally Posted by ignite

No!

What you did was just to check functions value at particular points.You have to prove in general.
Case I: You can easily see that as all
Case II: In this case and are positive while and are negative.But and
Case III: Here , ,

Your generalisation is almost correct,except for the degree part.Think about it again.

Yes i know now its only that way when the polynomial has even powers.

If you or anyone else can have a look at this,how do you sketch the curve: y= ((e^x)+x)/((e^-x)-x) ?

I found the intercept at x=0 but I don't know how to solve equations in the form of e^x=x.

And how do you solve the question uploaded in the picture?