Proof there is at least one root:

e^x = 3 - 2x

can you just do 0 = 3 - 2x

x = 1.5 ?

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- Apr 26th 2010, 02:41 AMDaniielProof 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, 03:05 AMpickslides
- Apr 26th 2010, 03:47 AMDaniiel
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, 06:10 AMGusbob
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, 07:33 AMPlato
- Apr 27th 2010, 12:11 AMDaniiel
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