show equation has one solution

• Dec 6th 2008, 10:22 AM
taha7
show equation has one solution
here is a question our prof gave us:

Show that the equation X + e^(X) - 5 = 0 has only one root in the interval [1,2].

i graphed the equation and saw that it is between 1 and 2

but i cant calculate it for sum reason. any ideas how? can u solve it using regular algebra? i dont thing wer allowed using calc but if thers no other way, i mite jus use calc if thers a way lol.

thanks.
• Dec 6th 2008, 01:28 PM
Gamma
Solution
Define $\displaystyle f(x)=x+e^x - 5$
Use the fact that this function is continuous over a connected space, so you have the intermediate value theorem. Check that $\displaystyle f(1)<0$ and $\displaystyle f(2)>0$ gives you that there exists some $\displaystyle x_o\in[1,2]$ for which $\displaystyle f(x_o)=0$. To show this point is unique, it suffices to show this function is strictly monotonic on this interval. How do you do this? I don't know if you have had calculus or not (if so, just take the derivative to see it is always positive: $\displaystyle f'(x)=1+e^x$is always positive), it is pretty clear this function is increasing everywhere by inspection since both $\displaystyle x$ and $\displaystyle e^x$ have this property and the sum of strictly monotonic functions must also be strictly monotonic. The -5 term at the end doesnt affect anything, why? by definition, for f to be strictly monotonic you need only show $\displaystyle x<y \Rightarrow f(x)<f(y)$ see what happens to the constant term?.