# no. of real solution in exp. equation

• Sep 26th 2013, 11:00 AM
jacks
no. of real solution in exp. equation
Total no. of real solution of $\displaystyle 2^x = 1-x^2$

My Trail:: Let $\displaystyle f(x) = 2^x$ and $\displaystyle g(x) = 1-x^2$

Now Using first derivative test, we will check whose slope is Increasing faster rate.

So $\displaystyle f{'}(x) = 2^x\cdot \ln (2)$ and $\displaystyle g^{'}(x) = -2x$

So for The $\displaystyle x\leq -1$, Slope of $\displaystyle f(x) >$ Slope of $\displaystyle g(x)$

So $\displaystyle f(x)$ is always above $\displaystyle g(x)$

So no solution for $\displaystyle x\leq -1$

and for $\displaystyle x \geq 0,$ Slope of $\displaystyle f(x) >$Slope of $\displaystyle g(x)$

So $\displaystyle f(x)$ is always above $\displaystyle g(x)$

So no solution for $\displaystyle x > 0$

Now we will calculate in $\displaystyle -1 < x < 0$

Now I Did not understand How can I find whether $\displaystyle f(x)$ is $\displaystyle g(x)$ or not

Thanks
• Sep 26th 2013, 04:54 PM
topsquark
Re: no. of real solution in exp. equation

And you missed an obvious solution: (x, y) = (0, 1).

-Dan
• Sep 26th 2013, 07:25 PM
jacks
Re: no. of real solution in exp. equation
Thanks moderator

Yes $\displaystyle x = 0$ is obivious solution of given equation.

Now my question is how can I check these two graph intersect in $\displaystyle -1< x < 0$ or not.

and How can we draw figure of these graph in $\displaystyle -1<x<0$

Thanks
• Sep 26th 2013, 09:32 PM
votan
Re: no. of real solution in exp. equation
Quote:

Originally Posted by jacks
Thanks moderator

Yes $\displaystyle x = 0$ is obivious solution of given equation.

Now my question is how can I check these two graph intersect in $\displaystyle -1< x < 0$ or not.

and How can we draw figure of these graph in $\displaystyle -1<x<0$

Thanks

The graph provided by Topskward is the answer to the last line of your post.
• Sep 26th 2013, 09:37 PM
chen09
Re: no. of real solution in exp. equation
Do you have to show proof (algebraic) ?
• Sep 26th 2013, 10:00 PM
votan
Re: no. of real solution in exp. equation
Quote:

Originally Posted by chen09
Do you have to show proof (algebraic) ?

It has to be iterative solution
• Sep 27th 2013, 07:37 AM
votan
Re: no. of real solution in exp. equation
Quote:

Originally Posted by votan
It has to be iterative solution

I tried Newton-Raphson. It either converges to 1 or it becomes unstable depending on initial guess. I think bisection method should work but I did not put time on it. I found the other root graphically: x = -0.57203.

2^(-0.57203) = 0.6727
1 - x^2 = 0.6728
• Sep 27th 2013, 12:45 PM
topsquark
Re: no. of real solution in exp. equation
Quote:

Originally Posted by votan
Topskward

!!!!!!

-Dan
• Sep 27th 2013, 03:42 PM
votan
Re: no. of real solution in exp. equation
Quote:

Originally Posted by topsquark
!!!!!!

-Dan

I was wandering what is this. I went back to may posts. I am still laughing from it. My apology for mispelling topsquark