Lösung 1.1:1a

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The derivative
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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>.
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<math>{f}'\left( -4 \right)</math>
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gives the function's instantaneous rate of change at the point
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<math>x=-\text{4}</math>, i.e. it is a measure of the function's value changes in the vicinity of
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<math>x=-\text{4}</math>.
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In the graph of the function, this derivative is equal to the slope of the tangent to the curve of function at the point
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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>.
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<math>x=-\text{4}</math>.
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{| align="center"
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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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[[Image:1_1_1_a1.gif|center]]
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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>
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<math>y=kx+m\text{ }</math>
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where
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<math>k={f}'\left( -\text{4} \right)</math>
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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
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<math>{f}'\left( -\text{4} \right)>0</math>.
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<math>f^{\,\prime}(-5) > 0\,</math>.
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At the point
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<math>x=\text{1}</math>, the tangent slopes downwards and this means that
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<math>{f}'\left( \text{1} \right)<0</math>.
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[[Image:1_1_1_a2.gif|center]]
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At the point <math>x=1</math>, the tangent slopes downwards and this means that
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<math>f^{\,\prime}(1) < 0\,</math>.
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<math>y=kx+m\text{ }</math>
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{| align="center"
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where
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||{{:1.1 - Figure - Solution - The graph of f(x) in exercise 1.1:1a with the tangent line at x = 1}}
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<math>k={f}'\left( 1 \right)</math>
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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>
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Version vom 08:48, 14. Okt. 2008

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\,.

1.1 - Figure - Solution - The graph of f(x) in exercise 1.1:1a with the tangent line at x = -5
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\,.

1.1 - Figure - Solution - The graph of f(x) in exercise 1.1:1a with the tangent line at x = 1
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“