Lösung 1.3:3e

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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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As always, a function can only have local extreme points at one of the following types of points,
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1. Critical points, i.e. where
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# critical points, i.e. where <math>f^{\,\prime}(x)=0</math>,
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<math>{f}'\left( x \right)=0</math>;
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# points where the function is not differentiable, and
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# endpoints of the interval of definition.
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2. Points where the function is not differentiable;
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3. Endpoints of the interval of definition.
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We investigate these three cases.
We investigate these three cases.
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1. We obtain the critical points by setting the derivative equal to zero:
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<ol>
 +
<li>We obtain the critical points by setting the derivative equal to zero,
 +
{{Displayed math||<math>\begin{align}
 +
f^{\,\prime}(x) &= (x^2-x-1)'e^x + (x^2-x-1)\bigl(e^x\bigr)^{\prime}\\[5pt]
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&= (2x-1)e^x + (x^2-x-1)e^x\\[5pt]
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&= (x^2+x-2)e^x\,\textrm{.}
 +
\end{align}</math>}}
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<math>\begin{align}
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This expression for the derivative can only be zero when <math>x^2+x-2=0</math>, because <math>e^x</math> differs from zero for all values of <math>x</math>. We solve the second-degree equation by completing the square,
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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} \\
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\end{align}</math>
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 +
{{Displayed math||<math>\begin{align}
 +
\Bigl(x+\frac{1}{2}\Bigr)^2 - \Bigl(\frac{1}{2}\Bigr)^2 - 2 &= 0\,,\\[5pt]
 +
\Bigl(x+\frac{1}{2}\Bigr)^2 &= \frac{9}{4}\,,\\[5pt]
 +
x+\frac{1}{2} &= \pm\frac{3}{2}\,,
 +
\end{align}</math>}}
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This expression for the derivative can only be zero when
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i.e. <math>x=-\tfrac{1}{2}-\tfrac{3}{2}=-2</math> and <math>x=-\tfrac{1}{2}+\tfrac{3}{2}=1</math>. Both of these points lie within the region of definition, <math>-3\le x\le 3\,</math>.</li>
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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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<li>The function is a polynomial <math>x^2-x-1</math> 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.</li>
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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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 +
<li>The function's region of definition is <math>-3\le x\le 3</math> and the endpoints <math>x=-3</math> and <math>x=3</math> are therefore possible local extreme points.</li>
 +
</ol>
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i.e.
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All in all, there are four points <math>x=-3</math>, <math>x=-2</math>, <math>x=1</math> and <math>x=3</math> where the function possibly has local extreme points.
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<math>x=-\frac{1}{2}-\frac{3}{2}=-2</math>
+
-
and
+
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<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>
+
 +
Now, we will write down a table of the sign of the derivative, in order to investigate if the function has local extreme points.
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2. The function is a polynomial
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We can factorize the derivative somewhat,
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<math>x^{2}-x-1</math>
+
-
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.
+
 +
{{Displayed math||<math>f^{\,\prime}(x) = (x^2+x-2)e^x = (x+2)(x-1)e^x\,,</math>}}
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3. The function's region of definition is
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since <math>x^2+x-2</math> 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:
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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
 
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<math>x=3</math>
 
-
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.
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{| border="1" cellpadding="5" cellspacing="0" align="center"
 +
|-
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|width="50px" align="center" style="background:#efefef;"| <math>x</math>
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|width="50px" align="center" style="background:#efefef;"| <math>-3</math>
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|style="background:#efefef;"|
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|width="50px" align="center" style="background:#efefef;"| <math>-2</math>
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|style="background:#efefef;"|
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|width="50px" align="center" style="background:#efefef;"| <math>1</math>
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|style="background:#efefef;"|
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|width="50px" align="center" style="background:#efefef;"| <math>3</math>
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|-
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|width="50px" align="center"| <math>x+2</math>
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|width="50px" align="center"| <math>-</math>
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|width="50px" align="center"| <math>-</math>
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|width="50px" align="center"| <math>0</math>
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|width="50px" align="center"| <math>+</math>
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|width="50px" align="center"| <math>+</math>
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|width="50px" align="center"| <math>+</math>
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|width="50px" align="center"| <math>+</math>
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|-
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|width="50px" align="center"| <math>x-1</math>
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|width="50px" align="center"| <math>-</math>
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|width="50px" align="center"| <math>-</math>
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|width="50px" align="center"| <math>-</math>
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|width="50px" align="center"| <math>-</math>
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|width="50px" align="center"| <math>0</math>
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|width="50px" align="center"| <math>+</math>
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|width="50px" align="center"| <math>+</math>
 +
|-
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|width="50px" align="center"| <math>e^x</math>
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|width="50px" align="center"| <math>+</math>
 +
|width="50px" align="center"| <math>+</math>
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|width="50px" align="center"| <math>+</math>
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|width="50px" align="center"| <math>+</math>
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|width="50px" align="center"| <math>+</math>
 +
|width="50px" align="center"| <math>+</math>
 +
|width="50px" align="center"| <math>+</math>
 +
|-
 +
|}
 +
 +
 +
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:
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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
 
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<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
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{| border="1" cellpadding="5" cellspacing="0" align="center"
 +
|-
 +
|width="50px" align="center" style="background:#efefef;"| <math>x</math>
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|width="50px" align="center" style="background:#efefef;"| <math>-3</math>
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|style="background:#efefef;"|
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|width="50px" align="center" style="background:#efefef;"| <math>-2</math>
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|style="background:#efefef;"|
 +
|width="50px" align="center" style="background:#efefef;"| <math>1</math>
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|style="background:#efefef;"|
 +
|width="50px" align="center" style="background:#efefef;"| <math>3</math>
 +
|-
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|width="50px" align="center"| <math>f^{\,\prime}(x)</math>
 +
|width="50px" align="center"| &nbsp;
 +
|width="50px" align="center"| <math>+</math>
 +
|width="50px" align="center"| <math>0</math>
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|width="50px" align="center"| <math>-</math>
 +
|width="50px" align="center"| <math>0</math>
 +
|width="50px" align="center"| <math>+</math>
 +
|width="50px" align="center"| &nbsp;
 +
|-
 +
|width="50px" align="center"| <math>f(x)</math>
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|width="50px" align="center"| <math>11e^{-3}</math>
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|width="50px" align="center"| <math>\nearrow</math>
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|width="50px" align="center"| <math>5e^{-2}</math>
 +
|width="50px" align="center"| <math>\searrow</math>
 +
|width="50px" align="center"| <math>-e</math>
 +
|width="50px" align="center"| <math>\nearrow</math>
 +
|width="50px" align="center"| <math>5e^3</math>
 +
|-
 +
|}
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The function has local minimum points at
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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>.
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<math>x=-3</math>
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and
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<math>x=1</math>, and local maximum points
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<math>x=-2</math>
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and
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<math>x=3</math>.
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Version vom 08:21, 21. 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 \displaystyle f^{\,\prime}(x)=0,
  2. points where the function is not differentiable, and
  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^{\,\prime}(x) &= (x^2-x-1)'e^x + (x^2-x-1)\bigl(e^x\bigr)^{\prime}\\[5pt] &= (2x-1)e^x + (x^2-x-1)e^x\\[5pt] &= (x^2+x-2)e^x\,\textrm{.} \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}

    \Bigl(x+\frac{1}{2}\Bigr)^2 - \Bigl(\frac{1}{2}\Bigr)^2 - 2 &= 0\,,\\[5pt] \Bigl(x+\frac{1}{2}\Bigr)^2 &= \frac{9}{4}\,,\\[5pt] x+\frac{1}{2} &= \pm\frac{3}{2}\,, \end{align}

    i.e. \displaystyle x=-\tfrac{1}{2}-\tfrac{3}{2}=-2 and \displaystyle x=-\tfrac{1}{2}+\tfrac{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 and \displaystyle x=3 are therefore possible local extreme points.

All in all, there are four points \displaystyle x=-3, \displaystyle x=-2, \displaystyle 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 if the function has local extreme points.

We can factorize the derivative somewhat,

\displaystyle f^{\,\prime}(x) = (x^2+x-2)e^x = (x+2)(x-1)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:


\displaystyle x \displaystyle -3 \displaystyle -2 \displaystyle 1 \displaystyle 3
\displaystyle x+2 \displaystyle - \displaystyle - \displaystyle 0 \displaystyle + \displaystyle + \displaystyle + \displaystyle +
\displaystyle x-1 \displaystyle - \displaystyle - \displaystyle - \displaystyle - \displaystyle 0 \displaystyle + \displaystyle +
\displaystyle e^x \displaystyle + \displaystyle + \displaystyle + \displaystyle + \displaystyle + \displaystyle + \displaystyle +


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:


\displaystyle x \displaystyle -3 \displaystyle -2 \displaystyle 1 \displaystyle 3
\displaystyle f^{\,\prime}(x)   \displaystyle + \displaystyle 0 \displaystyle - \displaystyle 0 \displaystyle +  
\displaystyle f(x) \displaystyle 11e^{-3} \displaystyle \nearrow \displaystyle 5e^{-2} \displaystyle \searrow \displaystyle -e \displaystyle \nearrow \displaystyle 5e^3


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.

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