Hey there!
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Remember what the function y=f'(x) actually denotes; it's just the slope of the function y=f(x). If f'(x)>0 over some domain, then f(x) is increasing over that domain. Similarly, if f'(x)<0 over some domain, then f(x) is decreasing over that domain. And if f'(x)=0, we have a stationary point.
Also, since
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\frac{d}{dx}f(x) = f'(x))
We have
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\int d(f(x)) = \int f'(x) dx \\ \int f'(x) dx = f(x) + C)
ie. the area under the curve of y=f'(x) denotes the function value of f(x) at a specific x value.
From my first point, we see that f'(x)>0 for 0≤x<2, hence it is for the same domain that f(x) is increasing.
From my second point, we see that the maximum signed area under the curve is 4 units. Hence, the maximum value of f(x) is also 4.
From my second point, we note that f(6) will be the total signed area from 0 to 6. Seeing that A
1 and A
2 cancel, we just note the area between x=4 and x=6 (which is just a rectangle!) to be 6 units squared; hence f(6)=-6 (the area is below the x-axis!).
The graph is rather more difficult.
Key points should be noted on the graph, and can be deduced from the signed area! For example, (0, 0), (2, 4), (4, 0) and (6, -6) should all be on your graph. Since f'(2)=0, there should be a stationary point there. Note that also the gradient of f'(x) is negative at x=2, this indicates that f''(2)<0 and thus we have a maximum at x=2. Also, ensure that the gradient at x=0 looks convincingly like a gradient of 3, and similarly for x=4, make sure the gradient looks convincingly like a gradient of -3. Since between x=4 and x=6 f'(x) is a constant, between x=4 and x=6 f(x) should also be a straight line that has a constant gradient of -3.
If you need an actual picture of the graph, let me know and I'll attach one.
Hope this helps
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