Aus Online Mathematik Brückenkurs 2

Version vom 14:37, 17. Okt. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 12:56, 10. Mär. 2009 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
Zeile 3:		Zeile 3:
	For this function, the derivative is given by		For this function, the derivative is given by
-	{{Displayed math||<math>f^{\,\prime}(x) = -3e^{-3x} + 5</math>}}	+	{{Abgesetzte Formel||<math>f^{\,\prime}(x) = -3e^{-3x} + 5</math>}}
	and if we set it to zero, we will obtain		and if we set it to zero, we will obtain
-	{{Displayed math||<math>3e^{-3x} = 5</math>}}	+	{{Abgesetzte Formel||<math>3e^{-3x} = 5</math>}}
	which is a first-degree equation in <math>e^{-3x}</math> and has the solution		which is a first-degree equation in <math>e^{-3x}</math> and has the solution
-	{{Displayed math||<math>x=-\frac{1}{3}\ln \frac{5}{3}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>x=-\frac{1}{3}\ln \frac{5}{3}\,\textrm{.}</math>}}
	The function therefore has a critical point <math>x=-\frac{1}{3}\ln \frac{5}{3}\,\textrm{.}</math>		The function therefore has a critical point <math>x=-\frac{1}{3}\ln \frac{5}{3}\,\textrm{.}</math>
Zeile 19:		Zeile 19:
	The second derivative is equal to		The second derivative is equal to
-	{{Displayed math||<math>f^{\,\prime\prime}(x) = -3\cdot (-3)e^{-3x} = 9e^{-3x}</math>}}	+	{{Abgesetzte Formel||<math>f^{\,\prime\prime}(x) = -3\cdot (-3)e^{-3x} = 9e^{-3x}</math>}}
	and is positive for all values of ''x'', since the exponential function is always positive.		and is positive for all values of ''x'', since the exponential function is always positive.
Zeile 25:		Zeile 25:
	In particular, this means that		In particular, this means that
-	{{Displayed math||<math>f^{\,\prime\prime}\Bigl( -\frac{1}{3}\ln \frac{5}{3} \Bigr) > 0\,,</math>}}	+	{{Abgesetzte Formel||<math>f^{\,\prime\prime}\Bigl( -\frac{1}{3}\ln \frac{5}{3} \Bigr) > 0\,,</math>}}
	which means that <math>x=-\tfrac{1}{3}\ln\tfrac{5}{3}</math> is a local minimum.		which means that <math>x=-\tfrac{1}{3}\ln\tfrac{5}{3}</math> is a local minimum.

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Version vom 12:56, 10. Mär. 2009

Because the function is defined and differentiable for all x, the function can only have local extreme points at the critical points, i.e. where the derivative is zero.

For this function, the derivative is given by

\displaystyle f^{\,\prime}(x) = -3e^{-3x} + 5

and if we set it to zero, we will obtain

\displaystyle 3e^{-3x} = 5

which is a first-degree equation in \displaystyle e^{-3x} and has the solution

\displaystyle x=-\frac{1}{3}\ln \frac{5}{3}\,\textrm{.}

The function therefore has a critical point \displaystyle x=-\frac{1}{3}\ln \frac{5}{3}\,\textrm{.}

Instead of studying the sign changes of the derivative around the critical point in order to decide if the point is a local maximum, minimum or neither, we investigate the function's second derivative.

The second derivative is equal to

\displaystyle f^{\,\prime\prime}(x) = -3\cdot (-3)e^{-3x} = 9e^{-3x}

and is positive for all values of x, since the exponential function is always positive.

In particular, this means that

\displaystyle f^{\,\prime\prime}\Bigl( -\frac{1}{3}\ln \frac{5}{3} \Bigr) > 0\,,

which means that \displaystyle x=-\tfrac{1}{3}\ln\tfrac{5}{3} is a local minimum.