Solution 1.1:1a

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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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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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{| 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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||<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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Because the tangent is sloping upwards, it has a positive gradient and therefore
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<math>f^{\,\prime}(-5) > 0\,</math>.
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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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{| 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 = 1}}
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|-
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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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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).
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