Lösung 1.3:3a

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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}
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{{Abgesetzte Formel||<math>\begin{align}
f^{\,\prime}(x)
f^{\,\prime}(x)
&= -4x^3 + 8\cdot 3x^2 - 18\cdot 2x\\[5pt]
&= -4x^3 + 8\cdot 3x^2 - 18\cdot 2x\\[5pt]
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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,
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,
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{{Displayed math||<math>x^2 - 6x + 9 = 0\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<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,
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{{Displayed math||<math>(x-3)^2 - 3^2 + 9 = 0</math>}}
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{{Abgesetzte Formel||<math>(x-3)^2 - 3^2 + 9 = 0</math>}}
which are, after simplifying,
which are, after simplifying,
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{{Displayed math||<math>(x-3)^2 = 0</math>}}
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{{Abgesetzte Formel||<math>(x-3)^2 = 0</math>}}
and this equation has the root <math>x=3</math>.
and this equation has the root <math>x=3</math>.
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Because the derivative can be written as
Because the derivative can be written as
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{{Displayed math||<math>f^{\,\prime}(x) = -4x(x-3)^2</math>}}
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{{Abgesetzte Formel||<math>f^{\,\prime}(x) = -4x(x-3)^2</math>}}
we start by writing down the sign changes for the factors <math>-4x</math> and
we start by writing down the sign changes for the factors <math>-4x</math> and

Version vom 12:55, 10. Mär. 2009

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).