Solution 1.1:1a

From Förberedande kurs i matematik 2

(Difference between revisions)
Jump to: navigation, search
Current revision (08:48, 14 October 2008) (edit) (undo)
m
 
Line 1: Line 1:
-
The derivative
+
The derivative <math>f^{\,\prime}(-5)</math> gives the function's instantaneous rate of change at the point <math>x=-5</math>, i.e. it is a measure of how much the function's value changes in the vicinity of <math>x=-5\,</math>.
-
<math>{f}'\left( -4 \right)</math>
+
-
gives the function's instantaneous rate of change at the point
+
-
<math>x=-\text{4}</math>, i.e. it is a measure of the function's value changes in the vicinity of
+
-
<math>x=-\text{4}</math>.
+
-
In the graph of the function, this derivative is equal to the slope of the tangent to the curve of function at the point
+
In the graph of the function, this derivative is equal to the slope of the tangent to the graph of the function at the point <math>x=-5\,</math>.
-
<math>x=-\text{4}</math>.
+
-
 
+
{| align="center"
-
 
+
||{{:1.1 - Figure - Solution - The graph of f(x) in exercise 1.1:1a with the tangent line at x = -5}}
-
+
|-
-
[[Image:1_1_1_a1.gif|center]]
+
||<small>The red tangent line has the equation<br>''y''&nbsp;=&nbsp;''kx''&nbsp;+&nbsp;''m'', where ''k''&nbsp;=&nbsp;''f'''(-5).</small>
-
 
+
|}
-
<math>y=kx+m\text{ }</math>
+
-
where
+
-
<math>k={f}'\left( -\text{4} \right)</math>
+
-
 
+
Because the tangent is sloping upwards, it has a positive gradient and therefore
Because the tangent is sloping upwards, it has a positive gradient and therefore
-
<math>{f}'\left( -\text{4} \right)>0</math>.
+
<math>f^{\,\prime}(-5) > 0\,</math>.
-
 
+
-
At the point
+
-
<math>x=\text{1}</math>, the tangent slopes downwards and this means that
+
-
<math>{f}'\left( \text{1} \right)<0</math>.
+
-
[[Image:1_1_1_a2.gif|center]]
+
At the point <math>x=1</math>, the tangent slopes downwards and this means that
 +
<math>f^{\,\prime}(1) < 0\,</math>.
-
<math>y=kx+m\text{ }</math>
+
{| align="center"
-
where
+
||{{:1.1 - Figure - Solution - The graph of f(x) in exercise 1.1:1a with the tangent line at x = 1}}
-
<math>k={f}'\left( 1 \right)</math>
+
|-
 +
||<small>The red tangent line has the equation<br>''y''&nbsp;=&nbsp;''kx''&nbsp;+&nbsp;''m'', where ''k''&nbsp;=&nbsp;''f'''(1).</small>
 +
|}

Current revision

The derivative \displaystyle f^{\,\prime}(-5) gives the function's instantaneous rate of change at the point \displaystyle x=-5, i.e. it is a measure of how much the function's value changes in the vicinity of \displaystyle x=-5\,.

In the graph of the function, this derivative is equal to the slope of the tangent to the graph of the function at the point \displaystyle x=-5\,.

[Image]

The red tangent line has the equation
y = kx + m, where k = f'(-5).

Because the tangent is sloping upwards, it has a positive gradient and therefore \displaystyle f^{\,\prime}(-5) > 0\,.

At the point \displaystyle x=1, the tangent slopes downwards and this means that \displaystyle f^{\,\prime}(1) < 0\,.

[Image]

The red tangent line has the equation
y = kx + m, where k = f'(1).
Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_1.1:1a"