Zeros and antiderivative

Problem

You are given the function \(f\) defined on \(\mathbb{R}\) with \(f(x)=e^x\cdot\left(2x+x^2\right)\).

1. Determine the zeros of the function \(f\).

2. Show that the function \(F\) defined on \(\mathbb{R}\) with \(F(x)=x^2\cdot e^x\) is an antiderivative of \(f\). Find another antiderivative \(G\) of \(f\) for which \(G(1)=2e\).

Solution of part a

As the exponential function always yields positive values, the zeros of \(f\) are given by the zeros of \(2x+x^2=x(2+x)\). By means of this factorization, the two zeros \(x_1=0\) und \(x_2=-2\) can easily be read off.

The result can also be reproduced with the help of Sage:

Solution of part b

In order to demonstrate that \(F\) is an antiderivative of \(f\), we need to prove that \(f\) is the derivative of \(F\). By means of the product rule we find

\[\frac{\mathrm{d}F}{\mathrm{d}x} = 2x\cdot\mathrm{e}^x+x^2\cdot\mathrm{e}^x =\mathrm{e}^x\cdot(2x+x^2) = f(x) .\]

This result can be verified by Sage as follows:

The second antiderivative \(G\) can differ from the antiderivative \(F\) only by a constant. It therefore is of the form

\[G(x)=x^2\cdot \mathrm{e}^x+c.\]

We use the requirement \(G(1)=2e\) to determine the constant \(c\), yielding \(c=\mathrm{e}\):