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.

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

**You are given m<0.**

Consider two regions:

$\displaystyle 0<x<\pi$:Here sin(x) is positive and mx is negative,hence they don't intersect.

$\displaystyle -\pi<x<0$:Here sin(x) is negative and mx is positive,hence they don't intersect.

Only intersection point is 0.

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

Quote:

Originally Posted by

**ignite** **You are given m<0.**

Consider two regions:

$\displaystyle 0<x<\pi$:Here sin(x) is positive and mx is negative,hence they don't intersect.

$\displaystyle -\pi<x<0$: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?

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

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.

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

Quote:

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:$\displaystyle x \ge 0 $ You can easily see that as all $\displaystyle 1,x,x^2,x^3,x^4 \ge 0 \Rightarrow F(x) > 0 $

Case II:$\displaystyle x \le -1 $ In this case $\displaystyle x^2 $ and $\displaystyle x^4$ are positive while $\displaystyle x $ and $\displaystyle x^3 $ are negative.But $\displaystyle x^2+x>0 $ and $\displaystyle x^4 + x^3 >0 \Rightarrow F(x) > 0$

Case III:$\displaystyle -1 < x < 0 $ Here $\displaystyle 1+x > 0 $ , $\displaystyle x^2 + x^3 >0 $ , $\displaystyle x^4 > 0 \Rightarrow F(x) > 0$

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

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

Quote:

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:$\displaystyle x \ge 0 $ You can easily see that as all $\displaystyle 1,x,x^2,x^3,x^4 \ge 0 \Rightarrow F(x) > 0 $

Case II:$\displaystyle x \le -1 $ In this case $\displaystyle x^2 $ and $\displaystyle x^4$ are positive while $\displaystyle x $ and $\displaystyle x^3 $ are negative.But $\displaystyle x^2+x>0 $ and $\displaystyle x^4 + x^3 >0 \Rightarrow F(x) > 0$

Case III:$\displaystyle -1 < x < 0 $ Here $\displaystyle 1+x > 0 $ , $\displaystyle x^2 + x^3 >0 $ , $\displaystyle x^4 > 0 \Rightarrow F(x) > 0$

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?Attachment 23721