Aus Online Mathematik Brückenkurs 2

As always, a function can only have local extreme points at one of the following types of points:

1. Critical points, i.e. where \displaystyle {f}'\left( x \right)=0;

2. Points where the function is not differentiable;

3. Endpoints of the interval of definition.

We investigate these three cases.

1. We obtain the critical points by setting the derivative equal to zero:

\displaystyle \begin{align} & {f}'\left( x \right)=\left( x^{2}-x-1 \right)^{\prime }e^{x}+\left( x^{2}-x-1 \right)\left( e^{x} \right)^{\prime } \\ & =\left( 2x-1 \right)e^{x}+\left( x^{2}-x-1 \right)e^{x} \\ & =\left( x^{2}+x-2 \right)e^{x} \\ \end{align}

This expression for the derivative can only be zero when \displaystyle x^{2}+x-2=0, because \displaystyle e^{x} differs from zero for all values of \displaystyle x. We solve the second-degree equation by completing the square:

\displaystyle \begin{align} & \left( x+\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}-2=0 \\ & =\left( x+\frac{1}{2} \right)^{2}=\frac{9}{4} \\ & =x+\frac{1}{2}=\pm \frac{3}{2} \\ \end{align}

i.e. \displaystyle x=-\frac{1}{2}-\frac{3}{2}=-2 and \displaystyle x=-\frac{1}{2}+\frac{3}{2}=1.Both of these points lie within the region of definition, \displaystyle -3\le x\le 3

2. The function is a polynomial \displaystyle x^{2}-x-1 multiplied by the exponential function \displaystyle e^{x}, and, because both of these functions are differentiable, the product is also a differentiable function, which shows that the function is differentiable everywhere.

3. The function's region of definition is \displaystyle -3\le x\le 3 and the endpoints \displaystyle x=-3\text{ } and \displaystyle x=3\text{ } are therefore possible local extreme points.

All in all, there are four points \displaystyle x=-3,\quad x=-2,\quad x=1 and \displaystyle x=3 where the function possibly has local extreme points.

Now, we will write down a table of the sign of the derivative, in order to investigate the function has local extreme points.

We can factorize the derivative somewhat,

\displaystyle {f}'\left( x \right)=\left( x^{2}+x-2 \right)e^{x}=\left( x+2 \right)\left( x-1 \right)e^{x}, since \displaystyle x^{2}+x-2 has zeros at \displaystyle x=-2 and \displaystyle x=1. Each individual factor in the derivative has a sign that is given in the table:

TABELL

The sign of the derivative is the product of these signs and from the derivative's sign we decide which local extreme points we have:

TABELL

The function has local minimum points at \displaystyle x=-3 and \displaystyle x=1, and local maximum points \displaystyle x=-2 and \displaystyle x=3.