Solution 1.1:1a

From Förberedande kurs i matematik 2

(Difference between revisions)
Jump to: navigation, search
m (Lösning 1.1:1a moved to Solution 1.1:1a: Robot: moved page)
Current revision (08:48, 14 October 2008) (edit) (undo)
m
 
(One intermediate revision not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
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>.
-
The derivative f'(-4)gives the function's instantaneous rate of change at the point x=-4, i.e. it is a measure of the function's value changes in the vicinity of x=-4.
+
-
In the graph of the function, this derivative is equal to the slope of the tangent to the curve of function at the point x=-4.
+
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>.
-
[[Image:1_1_1_a1.gif|center]]
+
{| align="center"
 +
||{{:1.1 - Figure - Solution - The graph of f(x) in exercise 1.1:1a with the tangent line at x = -5}}
 +
|-
 +
||<small>The red tangent line has the equation<br>''y''&nbsp;=&nbsp;''kx''&nbsp;+&nbsp;''m'', where ''k''&nbsp;=&nbsp;''f'''(-5).</small>
 +
|}
-
Because the tangent is sloping upwards, it has a positive gradient and therefore f'(-4)>0.
+
Because the tangent is sloping upwards, it has a positive gradient and therefore
 +
<math>f^{\,\prime}(-5) > 0\,</math>.
-
At the point x=1, the tangent slopes downwards and this means that f'(1)<0.
+
At the point <math>x=1</math>, the tangent slopes downwards and this means that
 +
<math>f^{\,\prime}(1) < 0\,</math>.
-
[[Image:1_1_1_a2.gif|center]]
+
{| align="center"
-
 
+
||{{:1.1 - Figure - Solution - The graph of f(x) in exercise 1.1:1a with the tangent line at x = 1}}
-
{{NAVCONTENT_STOP}}
+
|-
 +
||<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.math.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_1.1:1a"