HSC 2010 Question 9 Help (1 Viewer)

rand_althor · Jun 10, 2015

Hey could someone please explain (iii)?

InteGrand · Jun 10, 2015

rand_althor said:
Hey could someone please explain (iii)?

The total change of the function value from x = 0 to x = 4 is:

$\Delta f \equiv f(4) - f(0) = \int _0 ^4 f^\prime (x) \text{ d}x$ .

However, this integral is 0, as seen from the sketch.

So $f(4) - f(0) = 0 \Rightarrow f(4) = f(0) = 0$ .

From x = 4 to x = 6, the rate of change of the function value is just a constant -3 (since $f^\prime (x) = -3$ here), so the function is a straight line in this interval.

So the change in function value over this interval is:

$\text{change}\equiv f(6) - f(4) = (\text{rate of change})\times (\text{change in }x)\Rightarrow f(6) =\text{starting value (i.e. } f(4))+(\text{rate of change})\times (\text{change in }x)= f(4) + (-3)(6 - 4) = 0 -3\times 2 = -6$ .

HSC 2010 Question 9 Help (1 Viewer)

rand_althor

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InteGrand

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