# Show that f(x)=g(x) has a unique solution

• Jan 18th 2010, 12:29 PM
alpha
Show that f(x)=g(x) has a unique solution
i need help with this question please.

The functions f(x) and g(x) are defined below as:

f(x)=sinh(x)= [e^x- e^(-x)]/2 and g(x)= 1/(1+x^2 ) for all real x

Show that f(x)=g(x) has a unique solution for x in (-∞,∞).

Thanks for any help.
• Jan 18th 2010, 12:41 PM
Raoh
Quote:

Originally Posted by alpha
i need help with this question please.

The functions f(x) and g(x) are defined below as:

f(x)=sinh(x)= [e^x- e^(-x)]/2 and g(x)= 1/(1+x^2 ) for all real x

Show that f(x)=g(x) has a unique solution for x in (-∞,∞).

Thanks for any help.

i believe you should use the Intermediate value theorem.
• Jan 18th 2010, 12:45 PM
Jhevon
Quote:

Originally Posted by alpha
i need help with this question please.

The functions f(x) and g(x) are defined below as:

f(x)=sinh(x)= [e^x- e^(-x)]/2 and g(x)= 1/(1+x^2 ) for all real x

Show that f(x)=g(x) has a unique solution for x in (-∞,∞).

Thanks for any help.

Raoh is right. apply the theorem to the function h(x) = f(x) - g(x). show that h(x) = 0 for some x. (showing that h(x) = 0 is equivalent to showing f(x) = g(x), hope you see that) all you need to do is find two x-values where h(x) changes sign and apply the theorem (be sure to check all the hypotheses)
• Jan 18th 2010, 01:06 PM
alpha
i can only find one solution for x and it is 0.63351. Is that it?

thanks.
• Jan 18th 2010, 01:10 PM
Jhevon
Quote:

Originally Posted by alpha
i can only find one solution for x and it is 0.63351. Is that it?

thanks.

you were never asked to find a solution (your solution would only be an approximation, by the way), you were simply asked to show one exists. do this by applying the intermediate value theorem as i said and as you did in this thread.
• Jan 18th 2010, 01:14 PM
alpha
Quote:

Originally Posted by Jhevon
you were never asked to find a solution (your solution would only be an approximation, by the way), you were simply asked to show one exists. do this by applying the intermediate value theorem as i said and as you did in this thread.

how can i do that by using the values of positive and negative infinity though?
• Jan 18th 2010, 01:30 PM
Jhevon
Quote:

Originally Posted by alpha
how can i do that by using the values of positive and negative infinity though?

you don't have to use infinity. you can pick any two numbers you wish, as long as the functions change sign for those values. you can then apply the theorem on some closed interval containing those values.

if you really want to incorporate infinity, you'd have to use limits.
• Jan 18th 2010, 01:41 PM
alpha
i inserted 1 and -1 and it shows there is a sign change therefore the solution exists.
• Jan 18th 2010, 01:44 PM
Jhevon
Quote:

Originally Posted by alpha
i inserted 1 and -1 and it shows there is a sign change therefore the solution exists.

yes, or 0 and 1 would be nice. thus you can show a solution exists on the interval [0,1], which is included in the interval $\displaystyle (- \infty, \infty)$, hence, there is a solution on the latter interval, as you wanted to show.