Lösung 1.3:3a

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We need to investigate three types of points in order to determine the function's local extreme points,
We need to investigate three types of points in order to determine the function's local extreme points,
-
1. Critical points, i.e. where
+
# critical points, i.e. where <math>f^{\,\prime}(x) = 0</math>,
-
<math>{f}'\left( x \right)=0</math>;
+
# points where the function is not differentiable, and
 +
# endpoints of the interval of definition,
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2. Points where the function is not differentiable;
+
but we can ignore items 2 and 3, because the function is a polynomial and is therefore defined and differentiable everywhere.
-
 
+
-
3. Endpoints of the interval of definition.
+
-
 
+
-
but we can ignore 2. and 3., because the function is a polynomial and is therefore defined and differentiable everywhere.
+
The critical points are given by the points where the derivative,
The critical points are given by the points where the derivative,
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+
{{Displayed math||<math>\begin{align}
-
<math>\begin{align}
+
f^{\,\prime}(x)
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& {f}'\left( x \right)=-4x^{3}+8\centerdot 3x^{2}-18\centerdot 2x \\
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&= -4x^3 + 8\cdot 3x^2 - 18\cdot 2x\\[5pt]
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& =-4x^{3}+24x^{2}-36x \\
+
&= -4x^3 + 24x^2 - 36x\\[5pt]
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& =-4x\left( x^{2}-6x+9 \right) \\
+
&= -4x(x^2 - 6x + 9)
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\end{align}</math>
+
\end{align}</math>}}
-
 
+
is equal to zero.
is equal to zero.
-
From the factorized from of the derivative, we see that the derivative is zero when either the first factor,
+
From the factorized from of the derivative, we see that the derivative is zero when either the first factor, ''x'', is zero, or when the other factor is zero,
-
<math>x</math>, is zero, or when the other factor is zero:
+
-
 
+
-
 
+
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<math>x^{2}-6x+9=0</math>
+
 +
{{Displayed math||<math>x^2 - 6x + 9 = 0\,\textrm{.}</math>}}
We solve this second-degree equation by completing the square on the left-hand side,
We solve this second-degree equation by completing the square on the left-hand side,
 +
{{Displayed math||<math>(x-3)^2 - 3^2 + 9 = 0</math>}}
-
<math>\left( x-3 \right)^{2}-3^{2}+9=0</math>
+
which are, after simplifying,
 +
{{Displayed math||<math>(x-3)^2 = 0</math>}}
-
which are, after simplifying
+
and this equation has the root <math>x=3</math>.
 +
Therefore, the function has two critical points, <math>x=0</math> and <math>x=3</math>.
-
<math>\left( x-3 \right)^{2}=0</math>
+
The next step is to write down an outline of the derivative's sign changes, from which we then can see the possible local extreme points.
-
 
+
-
 
+
-
and this equation has the root
+
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<math>x=3</math>.
+
-
 
+
-
 
+
-
 
+
-
Therefore, the function has two critical points,
+
-
<math>x=0\text{ }</math>
+
-
and
+
-
<math>x=3</math>.
+
-
 
+
-
The next step is to write down an outline of the derivative's sign changes, from which we then can see the possible local extreme points.
+
Because the derivative can be written as
Because the derivative can be written as
 +
{{Displayed math||<math>f^{\,\prime}(x) = -4x(x-3)^2</math>}}
-
<math>{f}'\left( x \right)=-4x\left( x-3 \right)^{2}</math>
+
we start by writing down the sign changes for the factors <math>-4x</math> and
-
 
+
<math>(x-3)^{2}</math>.
-
 
+
-
we start by writing down the sign changes for the factors
+
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<math>-4x</math>
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-
and
+
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<math>\left( x-3 \right)^{2}</math>.
+
-
 
+
-
TABLE
+
-
 
+
-
The derivative, which is the product of these factors, has the sign changes given below, which are a consequence of the calculating rules for signs:
+
-
 
+
-
<math>+\centerdot +=+,\quad -\centerdot +=-,\quad \text{and}\quad -\centerdot -=+</math>
+
{| border="1" cellpadding="5" cellspacing="0" align="center"
 +
|-
 +
|width="50px" align="center" style="background:#efefef;"| <math>x</math>
 +
|width="50px" align="center" style="background:#efefef;"|
 +
|width="50px" align="center" style="background:#efefef;"| <math>0</math>
 +
|width="50px" align="center" style="background:#efefef;"|
 +
|width="50px" align="center" style="background:#efefef;"| <math>3</math>
 +
|width="50px" align="center" style="background:#efefef;"|
 +
|-
 +
|width="50px" align="center"| <math>-4x</math>
 +
|width="50px" align="center"| <math>+</math>
 +
|width="50px" align="center"| <math>0</math>
 +
|width="50px" align="center"| <math>-</math>
 +
|width="50px" align="center"| <math>-</math>
 +
|width="50px" align="center"| <math>-</math>
 +
|-
 +
|width="50px" align="center"| <math>(x-3)^2</math>
 +
|width="50px" align="center"| <math>+</math>
 +
|width="50px" align="center"| <math>+</math>
 +
|width="50px" align="center"| <math>+</math>
 +
|width="50px" align="center"| <math>0</math>
 +
|width="50px" align="center"| <math>+</math>
 +
|}
 +
The derivative, which is the product of these factors, has the sign changes given below, which are a consequence of the calculating rules for signs: <math>{+}\cdot {+}={+}</math>, <math>{-}\cdot {+} = {-}</math> and <math>{-}\cdot {-}={+}</math>.
-
TABLE
+
{| border="1" cellpadding="5" cellspacing="0" align="center"
 +
|-
 +
|width="50px" align="center" style="background:#efefef;"| <math>x</math>
 +
|width="50px" align="center" style="background:#efefef;"|
 +
|width="50px" align="center" style="background:#efefef;"| <math>0</math>
 +
|width="50px" align="center" style="background:#efefef;"|
 +
|width="50px" align="center" style="background:#efefef;"| <math>3</math>
 +
|width="50px" align="center" style="background:#efefef;"|
 +
|-
 +
|width="50px" align="center"| <math>f^{\,\prime}(x)</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"| <math>0</math>
 +
|width="50px" align="center"| <math>-</math>
 +
|-
 +
|width="50px" align="center"| <math>f(x)</math>
 +
|width="50px" align="center"| <math>\nearrow</math>
 +
|width="50px" align="center"| <math>0</math>
 +
|width="50px" align="center"| <math>\searrow</math>
 +
|width="50px" align="center"| <math>-27</math>
 +
|width="50px" align="center"| <math>\searrow</math>
 +
|}
-
From this, we see that
+
From this, we see that <math>x=0</math> is a local maximum, whilst <math>x=3</math>
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<math>x=0\text{ }</math>
+
-
is a local maximum, whilst
+
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<math>x=3</math>
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is an inflexion point (and therefore not an extreme point).
is an inflexion point (and therefore not an extreme point).

Version vom 13:42, 17. Okt. 2008

We need to investigate three types of points in order to determine the function's local extreme 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,

but we can ignore items 2 and 3, because the function is a polynomial and is therefore defined and differentiable everywhere.

The critical points are given by the points where the derivative,

\displaystyle \begin{align}

f^{\,\prime}(x) &= -4x^3 + 8\cdot 3x^2 - 18\cdot 2x\\[5pt] &= -4x^3 + 24x^2 - 36x\\[5pt] &= -4x(x^2 - 6x + 9) \end{align}

is equal to zero.

From the factorized from of the derivative, we see that the derivative is zero when either the first factor, x, is zero, or when the other factor is zero,

\displaystyle x^2 - 6x + 9 = 0\,\textrm{.}

We solve this second-degree equation by completing the square on the left-hand side,

\displaystyle (x-3)^2 - 3^2 + 9 = 0

which are, after simplifying,

\displaystyle (x-3)^2 = 0

and this equation has the root \displaystyle x=3.

Therefore, the function has two critical points, \displaystyle x=0 and \displaystyle x=3.

The next step is to write down an outline of the derivative's sign changes, from which we then can see the possible local extreme points.

Because the derivative can be written as

\displaystyle f^{\,\prime}(x) = -4x(x-3)^2

we start by writing down the sign changes for the factors \displaystyle -4x and \displaystyle (x-3)^{2}.

\displaystyle x \displaystyle 0 \displaystyle 3
\displaystyle -4x \displaystyle + \displaystyle 0 \displaystyle - \displaystyle - \displaystyle -
\displaystyle (x-3)^2 \displaystyle + \displaystyle + \displaystyle + \displaystyle 0 \displaystyle +

The derivative, which is the product of these factors, has the sign changes given below, which are a consequence of the calculating rules for signs: \displaystyle {+}\cdot {+}={+}, \displaystyle {-}\cdot {+} = {-} and \displaystyle {-}\cdot {-}={+}.

\displaystyle x \displaystyle 0 \displaystyle 3
\displaystyle f^{\,\prime}(x) \displaystyle + \displaystyle 0 \displaystyle - \displaystyle 0 \displaystyle -
\displaystyle f(x) \displaystyle \nearrow \displaystyle 0 \displaystyle \searrow \displaystyle -27 \displaystyle \searrow

From this, we see that \displaystyle x=0 is a local maximum, whilst \displaystyle x=3 is an inflexion point (and therefore not an extreme point).

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