Lösung 1.1:1a

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||{{:1.1 - Figure - Solution - The graph of f(x) in exercise 1.1:1a with the tangent line at x = -5}}
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||<small>The red tangent line has the equation<br>''y''&nbsp;=&nbsp;''kx''&nbsp;+&nbsp;''m'', where ''k''&nbsp;=&nbsp;''f'''(-5).</small>
||<small>The red tangent line has the equation<br>''y''&nbsp;=&nbsp;''kx''&nbsp;+&nbsp;''m'', where ''k''&nbsp;=&nbsp;''f'''(-5).</small>
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||<small>The red tangent line has the equation<br>''y''&nbsp;=&nbsp;''kx''&nbsp;+&nbsp;''m'', where ''k''&nbsp;=&nbsp;''f'''(1).</small>
||<small>The red tangent line has the equation<br>''y''&nbsp;=&nbsp;''kx''&nbsp;+&nbsp;''m'', where ''k''&nbsp;=&nbsp;''f'''(1).</small>
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Version vom 13:35, 10. Mär. 2009

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).
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.1:1a“