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# 2011 Calculus AB free response #4b

Absolute maximum over an interval. Critical points and differentiability. Created by Sal Khan.

## Want to join the conversation?

• why is the derivative f(x) and not f(t)
• The Fundamental Theorem of Calculus states that if you are trying to find the derivative of g(x)= definite integral from a to x of f(t)dt, then you simply replace the x for every t in the inner function. Therefore, g'(x) = f(x).
(1 vote)
• Is it possible to have a function that produce a circle and straight line?!
(1 vote)
• Not a circle, because it would fail the verticle line test, but anything less than or equal to a semicircle is allowed, so are straight lines. I can't think of one single way to define this function, but it could be defined piecewise.
• No, this questions does not allow calculators (someone confirm that I am right).
• Discuss the role of constant of integration in indefinite integrals, why we don't involve it in definite integrals? (Be precise)
• Maybe I can try it in a more general example. Let f(x)=d/dx F(x
The indefinite integral ∫ f(x) dx = F(x) + c
Now, if I want to find the definite integral (area under the curve) ∫ f(x) dx from a to b, the fundamental theorem tells us that this is the same thing as the indefinite integral evaluated at the endpoints. So we have ( F(b) + c ) - ( F(a) + c ). The constant of integration is there in both minuend and the subtrahend. Now, what happens if we distribute the subtraction?
F(b) + c - F(a) - c. We can simplify this! The two constants are the same, so they subtract to 0. Thus, our definite integral simplifies to F(b) - F(a). Notice: no constant. Because of the canceling out, we tend to ignore the constant of integration when taking definite integrals because it will always simply subtract out.
• the problem asks for the absolute maximum and the |-8 - 2pi| is clearly the largest so should the value be x = -4 and not x = 5/2
• No. Absolute maximum means the highest point on the graph of g(x), not the point where the absolute value of g(x) is the highest.
• I think Sal's answer is right, but I'm not sure about the evaluation at x = -4... I think 2 pi should have been added, not subtracted.
The formula for g(x) is 2x + area under f(x) from 0 to -4. There is clearly more positive area than negative area, so it should be 2x + some positive number (2 pi).
I'm thinking that since Sal didn't directly calculate the area, he forgot to check signs? Am I wrong?
(1 vote)
• It is subtracted because they reversed the normal way of expressing the interval which reversed the sign. (Integral from 0 to -4 instead of Integral from - 4 to 0) He explains this in the previous video.
• Why is the answer x=5/2, when the max on the graph is x=0?
(1 vote)
• at x=0, f(x) has a maximum, but we are looking for the maximum of g(x)
• what should i do to solve a problem for this video that i dont know with out watching the video agian or getting the problem wrong does any one have any tips for me
(1 vote)
• Post the problem and your solution and we can tell you whether you're correct.
• Hey Sal, I'm curious to know why you didn't g(3/2), where the straight line on the right side of this graph has a zero, would that be the absolute maximum? It seems to me that g(x) would be greater at that point than at g(5/2) because g(5/2) has negative area from integration lowering the number.
(1 vote)
• At g(3/2) = 2* (3/2) + integral(f(t),t,0,3/2) = 6/2 + 9/4 = 21/4 or 5.25
at Sal shows that g(5/2) = 6, which is larger.

It makes since if you remember that a function will increase as long as its derivative is positive and the function decreases when the derivative is negative. To find the points where a function changes from increasing to decreasing or decreasing to increasing, can generally be found by setting the first derivative to 0. For this function that only happened at x = 5/2
(1 vote)
• How do you know that it is a maximum? Couldn't it be a minimum?
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• As you know when solving f'(x) = 0 you obtain the horizontal slopes of the functions (which could be a maximun OR a minimun), another property that is not very known is that if you solve the SECOND derivative f''(x) you can find the concavity of the function, when positive the function is concave upward and thus for f'(x) = 0 for f''(x)>0 we have a minimun, same for the opposite => f'(x) = 0 AND f''(x) < 0 thus we have a maximun.

From: http://www.math.hmc.edu/calculus/tutorials/secondderiv/
(1 vote)