# Have I shown that one real number solution exists?

• Nov 22nd 2011, 04:58 PM
terrorsquid
Have I shown that one real number solution exists?
Show that there is exactly one real number \$\displaystyle x\$ for which\$\displaystyle e^x + x = 5\$. Find the integer part of this number. State any theorems you use in your solution.

My attempt:

\$\displaystyle e^x+x-5 = 0\$

\$\displaystyle f(0) = 1-5 = -4\$

\$\displaystyle f(2) \approx 7+2-5 = 4\$

By the intermediate value theorem, the continuous function crosses the x axis. The function is always increasing, therefore it crosses the x axis only once.

Is there an algebraic way for finding the integer part? I just subbed 1 into the original equation and got an answer < 5 then subbed in 2 and got an answer > 5 and deduced that x must be 1.something \$\displaystyle \therefore\$ the integer part is 1.
• Nov 22nd 2011, 05:10 PM
Prove It
Re: Have I shown that one real number solution exists?
Quote:

Originally Posted by terrorsquid
Show that there is exactly one real number \$\displaystyle x\$ for which\$\displaystyle e^x + x = 5\$. Find the integer part of this number. State any theorems you use in your solution.

My attempt:

\$\displaystyle e^x+x-5 = 0\$

\$\displaystyle f(0) = 1-5 = -4\$

\$\displaystyle f(2) \approx 7+2-5 = 4\$

By the intermediate value theorem, the continuous function crosses the x axis. The function is always increasing, therefore it crosses the x axis only once.

Is there an algebraic way for finding the integer part? I just subbed 1 into the original equation and got an answer < 5 then subbed in 2 and got an answer > 5 and deduced that x must be 1.something \$\displaystyle \therefore\$ the integer part is 1.

You haven't shown yet that there's EXACTLY one solution though...
• Nov 22nd 2011, 05:11 PM
TKHunny
Re: Have I shown that one real number solution exists?
Where did you demonstrate "only increasing"?

Define "algebraic". Iterative techniques are required.
• Nov 22nd 2011, 05:21 PM
terrorsquid
Re: Have I shown that one real number solution exists?
\$\displaystyle f'(x) = e^x + 1\$ which is positive for any value of x, therefore it is always increasing. Is that sufficient?

Originally I just noticed that the function was basically a shifted version of the e^x function and deemed it as always increasing :D Not very thorough.
• Nov 22nd 2011, 05:23 PM
Deveno
Re: Have I shown that one real number solution exists?
he's left out some steps, but if you accept that the function is always increasing, then there must be exactly one zero.

@terrorsquid: it would be prudent to show that f'(x) > 0, for all x (EDIT: i didn't see your latest post).