# Proof there is a root

• Apr 26th 2010, 03:41 AM
Daniiel
Proof there is a root
Proof there is at least one root:
e^x = 3 - 2x

can you just do 0 = 3 - 2x
x = 1.5 ?
• Apr 26th 2010, 04:05 AM
pickslides
Quote:

Originally Posted by Daniiel

can you just do 0 = 3 - 2x
x = 1.5 ?

No you can't, you could graph both $e^x$ and $3 - 2x$

If they intersect you have proven there is a solution.
• Apr 26th 2010, 04:47 AM
Daniiel
Thanks
http://img594.imageshack.us/img594/8...444ib81340.gif
Do you think thats all they want?
Not like a written out proof and a value for the intercept?
• Apr 26th 2010, 07:10 AM
Gusbob
Quote:

Originally Posted by Daniiel
Thanks
http://img594.imageshack.us/img594/8...444ib81340.gif
Do you think thats all they want?
Not like a written out proof and a value for the intercept?

If they don't specify what method to use, you should be able to use graphing. You should definitely do this unless you're doing a introductory calculus course. If you are doing an intro calculus course, they may want you to consider the following:

You can prove a solution exists without graphing, but you won't be able to find the exact value of the solution.

I won't go into detail here because the graphing should suffice. Basically note that e^x is always increasing (look at the derivative to see why) and 3 -2x is always decreasing. Since they both exist for all real x, and the range for y = 3 - 2x is all real values of y, they must intersect at some point. To write it out proper you may have to look into the intermediate value theorem.
• Apr 26th 2010, 08:33 AM
Plato
Quote:

Originally Posted by Daniiel
Proof there is at least one root:
e^x = 3 - 2x

Let $f(x)=e^x+2x-3$
Then $f(0)=-2<0~\&~f(1)-e+2-3>0$.
Therefore there is a root between 0 & 1.
• Apr 27th 2010, 01:11 AM
Daniiel
Thanks
i'll do both
i didn't think they're give us another IV thrm question because we got one in the last assignment and i thought you might have to prove its continuous aswell
Thanks again