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Lösung 1.3:3e

Aus Online Mathematik Brückenkurs 2

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K (Lösning 1.3:3e moved to Solution 1.3:3e: Robot: moved page)
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As always, a function can only have local extreme points at one of the following types of points:
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<center> [[Image:1_3_3e-1(4).gif]] </center>
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1. Critical points, i.e. where
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<math>{f}'\left( x \right)=0</math>;
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<center> [[Image:1_3_3e-2(4).gif]] </center>
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2. Points where the function is not differentiable;
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<center> [[Image:1_3_3e-3(4).gif]] </center>
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3. Endpoints of the interval of definition.
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We investigate these three cases.
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<center> [[Image:1_3_3e-4(4).gif]] </center>
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1. We obtain the critical points by setting the derivative equal to zero:
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 +
 
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<math>\begin{align}
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& {f}'\left( x \right)=\left( x^{2}-x-1 \right)^{\prime }e^{x}+\left( x^{2}-x-1 \right)\left( e^{x} \right)^{\prime } \\
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& =\left( 2x-1 \right)e^{x}+\left( x^{2}-x-1 \right)e^{x} \\
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& =\left( x^{2}+x-2 \right)e^{x} \\
 +
\end{align}</math>
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 +
 
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This expression for the derivative can only be zero when
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<math>x^{2}+x-2=0</math>, because
 +
<math>e^{x}</math>
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differs from zero for all values of
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<math>x</math>.
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We solve the second-degree equation by completing the square:
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 +
<math>\begin{align}
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& \left( x+\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}-2=0 \\
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& =\left( x+\frac{1}{2} \right)^{2}=\frac{9}{4} \\
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& =x+\frac{1}{2}=\pm \frac{3}{2} \\
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\end{align}</math>
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 +
 
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i.e.
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<math>x=-\frac{1}{2}-\frac{3}{2}=-2</math>
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and
 +
<math>x=-\frac{1}{2}+\frac{3}{2}=1</math>.Both of these points lie within the region of definition,
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<math>-3\le x\le 3</math>
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 +
 
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2. The function is a polynomial
 +
<math>x^{2}-x-1</math>
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multiplied by the exponential function
 +
<math>e^{x}</math>, and, because both of these functions are differentiable, the product is also a differentiable function, which shows that the function is differentiable everywhere.
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 +
 
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3. The function's region of definition is
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<math>-3\le x\le 3</math>
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and the endpoints
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<math>x=-3\text{ }</math>
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and
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<math>x=3\text{ }</math>
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are therefore possible local extreme points.
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All in all, there are four points
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<math>x=-3,\quad x=-2,\quad x=1</math>
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and
 +
<math>x=3</math>
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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,
 +
 
 +
<math>{f}'\left( x \right)=\left( x^{2}+x-2 \right)e^{x}=\left( x+2 \right)\left( x-1 \right)e^{x}</math>,
 +
since
 +
<math>x^{2}+x-2</math>
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has zeros at
 +
<math>x=-2</math>
 +
and
 +
<math>x=1</math>. 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
 +
<math>x=-3</math>
 +
and
 +
<math>x=1</math>, and local maximum points
 +
<math>x=-2</math>
 +
and
 +
<math>x=3</math>.

Version vom 13:21, 15. Okt. 2008

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

1. Critical points, i.e. where fx=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:


fx=x2−x−1ex+x2−x−1ex=2x−1ex+x2−x−1ex=x2+x−2ex


This expression for the derivative can only be zero when x2+x−2=0, because ex differs from zero for all values of x. We solve the second-degree equation by completing the square:

x+212−212−2=0=x+212=49=x+21=23


i.e. x=−21−23=−2 and x=−21+23=1.Both of these points lie within the region of definition, −3x3


2. The function is a polynomial x2−x−1 multiplied by the exponential function ex, 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 −3x3 and the endpoints x=−3 and x=3 are therefore possible local extreme points.

All in all, there are four points x=−3x=−2x=1 and 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,

fx=x2+x−2ex=x+2x−1ex , since x2+x−2 has zeros at x=−2 and 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 x=−3 and x=1, and local maximum points x=−2 and x=3.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.3:3e“