Aus Online Mathematik Brückenkurs 2

Version vom 14:32, 16. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 1.3:3e moved to Solution 1.3:3e: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 13:21, 15. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	As always, a function can only have local extreme points at one of the following types of points:
-	<center> [[Image:1_3_3e-1(4).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	1. Critical points, i.e. where
-	{{NAVCONTENT_START}}	+	<math>{f}'\left( x \right)=0</math>;
-	<center> [[Image:1_3_3e-2(4).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	2. Points where the function is not differentiable;
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:1_3_3e-3(4).gif]] </center>	+	3. Endpoints of the interval of definition.
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+	We investigate these three cases.
-	<center> [[Image:1_3_3e-4(4).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	1. We obtain the critical points by setting the derivative equal to zero:
		+
		+
		+	<math>\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}</math>
		+
		+
		+	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:
		+
		+	<math>\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}</math>
		+
		+
		+	i.e.
		+	<math>x=-\frac{1}{2}-\frac{3}{2}=-2</math>
		+	and
		+	<math>x=-\frac{1}{2}+\frac{3}{2}=1</math>.Both of these points lie within the region of definition,
		+	<math>-3\le x\le 3</math>
		+
		+
		+	2. 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.
		+
		+
		+	3. The function's region of definition is
		+	<math>-3\le x\le 3</math>
		+	and the endpoints
		+	<math>x=-3\text{ }</math>
		+	and
		+	<math>x=3\text{ }</math>
		+	are therefore possible local extreme points.
		+
		+	All in all, there are four points
		+	<math>x=-3,\quad x=-2,\quad x=1</math>
		+	and
		+	<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.
		+
		+	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>
		+	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 \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.