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nit

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The general idea is right...it looks a little wonky and asymmetric, that's all
 

za

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look at the left side of the origin. the gradient is -ve. doesnt that mean that the graph of f(x) should be decreasing?
 

za

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nah mate i think its the other way around. the blue line is f ' (x) and the red f(x)
 

za

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look at the graph.
wen the blue line crosses the x-axis there are turning points on the red one.
 

CM_Tutor

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Sorry, but in the graph on page 102, bottom left, the blue is the function f(x) and the red is its derivative f'(x). The y-intercept of the blue is a TP, thus its derivative (red) is zero. As the blue goes to +∞, it flattens, and its gradient goes to zero. Thus, red, f'(x), goes to zero from above. Same for goes for -∞ - f'(x) goes to zero from below.

Now, consider the other interpretation. Red = f(x), Blue = f'(x). At x = 0, Red is increasing, hence it has a positive slope. Its derivative would thus be positive, but blue is negative. Furthermore, as x goes to +∞ red decrease towards zero, and so its derivative is negative and approaching zero from below. This is not what blue is doing.
 
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