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

• Apr 26th 2012, 10:26 AM
syedmaheen
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.
• Apr 26th 2012, 11:14 AM
ignite
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.
• Apr 26th 2012, 01:08 PM
syedmaheen
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?
• Apr 26th 2012, 01:11 PM
syedmaheen
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.
• Apr 26th 2012, 01:58 PM
ignite
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.
• Apr 27th 2012, 01:38 AM
syedmaheen
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