From Förberedande kurs i matematik 2

Revision as of 14:32, 16 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 1.3:3c moved to Solution 1.3:3c: Robot: moved page) ← Previous diff		Current revision (07:28, 20 October 2008) (edit) (undo) Tek (Talk | contribs) m
(One intermediate revision not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	The only points which can possibly be local extreme points of the function are one of the following,
-	<center> [[Image:1_3_3c-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	# critical points, i.e. where <math>f^{\,\prime}(x) = 0\,</math>,
-	{{NAVCONTENT_START}}	+	# points where the function is not differentiable, and
-	<center> [[Image:1_3_3c-2(2).gif]] </center>	+	# endpoints of the interval of definition.
-	{{NAVCONTENT_STOP}}	+
		+	What determines the function's region of definition is <math>\ln x</math>, which is defined for <math>x > 0</math>, and this region does not have any endpoints (<math>x=0</math> does not satisfy <math>x>0</math>), so item 3 above does not give rise to any imaginable extreme points. Furthermore, the function is differentiable everywhere (where it is defined), because it consists of <math>x</math> and <math>\ln x</math> which are differentiable functions; so, item 2 above does not contribute any extreme points either.
		+
		+	All the remains are possibly critical points. We differentiate the function
		+
		+	{{Displayed math||<math>f^{\,\prime}(x) = 1\cdot \ln x + x\cdot \frac{1}{x} - 0 = \ln x+1</math>}}
		+
		+	and see that the derivative is zero when
		+
		+	{{Displayed math||<math>\ln x = -1\quad \Leftrightarrow \quad x = e^{-1}\,\textrm{.}</math>}}
		+
		+	In order to determine whether this is a local maximum, minimum or saddle point, we calculate the second derivative, <math>f^{\,\prime\prime}(x) = 1/x</math>, which gives that
		+
		+	{{Displayed math||<math>f^{\,\prime\prime}\bigl(e^{-1}\bigr) = \frac{1}{e^{-1}} = e > 0\,,</math>}}
		+
		+	which implies that <math>x=e^{-1}</math> is a local minimum.

---

Current revision

The only points which can possibly be local extreme points of the function are one of the following,

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.

What determines the function's region of definition is \displaystyle \ln x, which is defined for \displaystyle x > 0, and this region does not have any endpoints (\displaystyle x=0 does not satisfy \displaystyle x>0), so item 3 above does not give rise to any imaginable extreme points. Furthermore, the function is differentiable everywhere (where it is defined), because it consists of \displaystyle x and \displaystyle \ln x which are differentiable functions; so, item 2 above does not contribute any extreme points either.

All the remains are possibly critical points. We differentiate the function

\displaystyle f^{\,\prime}(x) = 1\cdot \ln x + x\cdot \frac{1}{x} - 0 = \ln x+1

and see that the derivative is zero when

\displaystyle \ln x = -1\quad \Leftrightarrow \quad x = e^{-1}\,\textrm{.}

In order to determine whether this is a local maximum, minimum or saddle point, we calculate the second derivative, \displaystyle f^{\,\prime\prime}(x) = 1/x, which gives that

\displaystyle f^{\,\prime\prime}\bigl(e^{-1}\bigr) = \frac{1}{e^{-1}} = e > 0\,,

which implies that \displaystyle x=e^{-1} is a local minimum.