Solution 1.3:2c

From Förberedande kurs i matematik 2

(Difference between revisions)
Jump to: navigation, search
Current revision (12:59, 17 October 2008) (edit) (undo)
m
 
(3 intermediate revisions not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
There are three types of points at which the function can have local extreme points,
-
<center> [[Bild:1_3_2c-1(3).gif]] </center>
+
-
{{NAVCONTENT_STOP}}
+
-
{{NAVCONTENT_START}}
+
-
<center> [[Bild:1_3_2c-2(3).gif]] </center>
+
-
{{NAVCONTENT_STOP}}
+
-
{{NAVCONTENT_START}}
+
-
<center> [[Bild:1_3_2c-3(3).gif]] </center>
+
-
{{NAVCONTENT_STOP}}
+
-
[[Bild:1_3_2_c.gif|center]]
+
# critical points, i.e. where <math>f^{\,\prime}(x)=0</math>,
 +
# points where the function is not differentiable, and
 +
# endpoints of the interval of definition.
 +
 
 +
Because our function is a polynomial, it is defined and differentiable everywhere, and therefore does not have any points which satisfy items 2 and 3.
 +
 
 +
As regards item 1, we set the derivative equal to zero and obtain the equation
 +
 
 +
{{Displayed math||<math>f^{\,\prime}(x) = 6x^2+6x-12 = 0\,\textrm{.}</math>}}
 +
 
 +
Dividing both sides by 6 and completing the square, we obtain
 +
 
 +
{{Displayed math||<math>\Bigl(x+\frac{1}{2}\Bigr)^2 - \Bigl(\frac{1}{2}\Bigr)^2 - 2 = 0\,\textrm{.}</math>}}
 +
 
 +
This gives us the equation
 +
 
 +
{{Displayed math||<math>\Bigl(x+\frac{1}{2}\Bigr)^2 = \frac{9}{4}</math>}}
 +
 
 +
and taking the square root gives the solutions
 +
 
 +
{{Displayed math||<math>\begin{align}
 +
x &= -\frac{1}{2}-\sqrt{\frac{9}{4}} = -\frac{1}{2}-\frac{3}{2} = -2\,,\\[5pt]
 +
x &= -\frac{1}{2}+\sqrt{\frac{9}{4}} = -\frac{1}{2}+\frac{3}{2} = 1\,\textrm{.}
 +
\end{align}</math>}}
 +
 
 +
This means that if the function has several extreme points, they must be among
 +
<math>x=-2</math> and <math>x=1</math>.
 +
 
 +
Then, we write down a sign table for the derivative, and read off the possible extreme points.
 +
 
 +
{| 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>-2</math>
 +
|width="50px" align="center" style="background:#efefef;"|
 +
|width="50px" align="center" style="background:#efefef;"| <math>1</math>
 +
|width="50px" align="center" style="background:#efefef;"|
 +
|-
 +
|width="50px" align="center"| <math>f^{\,\prime}(x)</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>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>21</math>
 +
|width="50px" align="center"| <math>\searrow</math>
 +
|width="50px" align="center"| <math>-6</math>
 +
|width="50px" align="center"| <math>\nearrow</math>
 +
|}
 +
 
 +
 
 +
The function has a local maximum at <math>x=-2</math> and a local minimum at <math>x=1</math>.
 +
 
 +
We obtain the overall appearance of the graph of the function from the table and by calculating the value of the function at a few points.
 +
 
 +
[[Image:1_3_2_c.gif|center]]

Current revision

There are three types of points at which the function can have 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.

Because our function is a polynomial, it is defined and differentiable everywhere, and therefore does not have any points which satisfy items 2 and 3.

As regards item 1, we set the derivative equal to zero and obtain the equation

\displaystyle f^{\,\prime}(x) = 6x^2+6x-12 = 0\,\textrm{.}

Dividing both sides by 6 and completing the square, we obtain

\displaystyle \Bigl(x+\frac{1}{2}\Bigr)^2 - \Bigl(\frac{1}{2}\Bigr)^2 - 2 = 0\,\textrm{.}

This gives us the equation

\displaystyle \Bigl(x+\frac{1}{2}\Bigr)^2 = \frac{9}{4}

and taking the square root gives the solutions

\displaystyle \begin{align}

x &= -\frac{1}{2}-\sqrt{\frac{9}{4}} = -\frac{1}{2}-\frac{3}{2} = -2\,,\\[5pt] x &= -\frac{1}{2}+\sqrt{\frac{9}{4}} = -\frac{1}{2}+\frac{3}{2} = 1\,\textrm{.} \end{align}

This means that if the function has several extreme points, they must be among \displaystyle x=-2 and \displaystyle x=1.

Then, we write down a sign table for the derivative, and read off the possible extreme points.

\displaystyle x \displaystyle -2 \displaystyle 1
\displaystyle f^{\,\prime}(x) \displaystyle + \displaystyle 0 \displaystyle - \displaystyle 0 \displaystyle +
\displaystyle f(x) \displaystyle \nearrow \displaystyle 21 \displaystyle \searrow \displaystyle -6 \displaystyle \nearrow


The function has a local maximum at \displaystyle x=-2 and a local minimum at \displaystyle x=1.

We obtain the overall appearance of the graph of the function from the table and by calculating the value of the function at a few points.

Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_1.3:2c"