If f'(x) is positive, then f(x) is increasing
If f'(x) is negative, then f(x) is decreasing.
We note that f'(x) is positive for
, then negative for all
This means f(x) increases until x=2, and then decreases. Note that f'(2)=0 which means at x=2, f(x) does have a stationary point.
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So we know that the maximum value is at x=2 using the above info. But then what is the value itself.
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If this doesn't make sense, be worried:
The area under the curve y=f'(x) between a and b is
 dx } \\ =f\left( b \right) -f\left( a \right) )
So the area under the curve y=f'(x) between x=0 and x=2 is
 dx } \\ =f\left( 2 \right) -f\left( 0 \right) )
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Now, if the y-coordinate at x=2 is f(2)
And the y-coordinate at x=0 is f(0)
Then the DIFFERENCE between the y-coordinates is also f(2)-f(0)
This means, that the area under the curve in the derivative, is the change in y-value of the original function.
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This means, that if f(0) = 0 like the question said, and the area under the derivative between 0 and 2 is A1 = 4, then f(x) increases by 4 units from x=0 to x=2. So the maximum value is 4
 &= f(4) - 3\times (6-4) \quad (\text{as the change in }x \text{ is } (6-4)) \\ &= 0-3\times 2 \\ &= -6. \end{align*}$$)