TITLE: Help me in showing that there is only 2 solutions for $f(x)=(x-1)e^{2x}+x^2-x-1=0$?
QUESTION [3 upvotes]: I have a problem with part (b) in the following question
Question
a) Function: $f(x)=(x-1)e^{2x}+x^2-x-1$, show that $f(x)$ has a zero between −1 and 1/2 and between 1/2 and 2.
b) Show that there are only exactly two solutions (two 0's).

My Approach
a) Pretty easy:
$$f(x)=(x-1)e^{2x}+x^2-x-1=0$$
$f(-1)>0,f(1/2)<0$ Intermediate Theorem tells that there is a solution.
$f(1/2)<0,f(2)>0$ Intermediate Theorem tells that there is a solution in between.
b) This is where I have the problem with, how do I show exactly that there is only 2 solutions. The most obvious thing that came to my mind is that the highest power is 2. But then I thought it can't be that easy, there is something I am missing.

REPLY [0 votes]: Try drawing rough curves for $e^{2x}=-x+1/(x-1)$.
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