# Thread: Proof there is a root

1. ## 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 ?

2. Originally Posted by Daniiel

can you just do 0 = 3 - 2x
x = 1.5 ?
No you can't, you could graph both $\displaystyle e^x$ and $\displaystyle 3 - 2x$

If they intersect you have proven there is a solution.

3. Thanks

Do you think thats all they want?
Not like a written out proof and a value for the intercept?

4. Originally Posted by Daniiel
Thanks

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.

5. Originally Posted by Daniiel
Proof there is at least one root:
e^x = 3 - 2x
Let $\displaystyle f(x)=e^x+2x-3$
Then $\displaystyle f(0)=-2<0~\&~f(1)-e+2-3>0$.
Therefore there is a root between 0 & 1.

6. 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