zeesbrat3
  • zeesbrat3
Which of the following statements is/are true? a.) If f '(c) = 0, then f has a local maximum or minimum at x = c. b.) If f is continuous on [a, b] and differentiable on (a, b) and f '(x) = 0 on (a, b), then f is constant on [a, b]. c.) The Mean Value Theorem can be applied to f(x) = 1/x2 on the interval [−1, 1].
Mathematics
chestercat
  • chestercat
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zeesbrat3
  • zeesbrat3
Please help
Michele_Laino
  • Michele_Laino
hint: for last statement, we have that x=0 belongs to interval [-1,1], and at x=0, the functio f(x)= 1/x^2 is not continuous

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zeesbrat3
  • zeesbrat3
Because it wouldnt exist?
Michele_Laino
  • Michele_Laino
since we can not divide by zero
zeesbrat3
  • zeesbrat3
Sorry, I thought of limits when I said it doesn't exist since like you said, you can't divide by 0
zeesbrat3
  • zeesbrat3
I think A i incorrect as well
Michele_Laino
  • Michele_Laino
yes! I think so, since we can have this case: |dw:1437332734543:dw|
zeesbrat3
  • zeesbrat3
I did it algebraically, but got the same thing. I figured if you had f(x) = x^3 then you wouldn't get a max or min anyways, just a hip
zeesbrat3
  • zeesbrat3
So, since A and C are incorrect, the only valid answer would be B, yes?
Michele_Laino
  • Michele_Laino
another counterexample is given by this function: f(x)= x^3 we have: f '(x) = 3x^2 and f '(0)==, nevertheless x=0 is not a point of local maximum or minimum |dw:1437333090875:dw|
Michele_Laino
  • Michele_Laino
oops.. f '(0)=0
Michele_Laino
  • Michele_Laino
yes! option B is the correct one
zeesbrat3
  • zeesbrat3
Thank you for clarifying C for me. :) Would you mind helping me with another?
Michele_Laino
  • Michele_Laino
ok!
zeesbrat3
  • zeesbrat3
Given f '(x) = (x − 4)(4 − 2x), find the x-coordinate for the relative maximum on the graph of f(x).
Michele_Laino
  • Michele_Laino
In general when I have to solve a problem like this, I develop the multiplications, namely I rewrite your function as follows: \[f\left( x \right) = - 2{x^2} + 12x - 16\]
Michele_Laino
  • Michele_Laino
then I solve this inequality: \[ - 2{x^2} + 12x - 16 \geqslant 0\]
Michele_Laino
  • Michele_Laino
which is equivalent to this one: \[{x^2} - 6x + 8 \leqslant 0\]
zeesbrat3
  • zeesbrat3
I got \[-2(x-4)(x-2)\ge0\]
Michele_Laino
  • Michele_Laino
yes! it is right!
zeesbrat3
  • zeesbrat3
Then divide both sides by -2. So 4 is the max?
zeesbrat3
  • zeesbrat3
Because it is the highest point
Michele_Laino
  • Michele_Laino
we have to proceed with caution
Michele_Laino
  • Michele_Laino
the solution of our inequlity is: \[2 \leqslant x \leqslant 4\]
Michele_Laino
  • Michele_Laino
inequality*
Michele_Laino
  • Michele_Laino
so, we have thi drawing: |dw:1437333653697:dw|
Michele_Laino
  • Michele_Laino
where f '(x) is positive then f(x) is increasing where f '(x ) is negative then f(x) is decreasing
zeesbrat3
  • zeesbrat3
The number line is your checks to see whether it is min or max..
Michele_Laino
  • Michele_Laino
yes!
Michele_Laino
  • Michele_Laino
|dw:1437333827000:dw|
perl
  • perl
Given f '(x) = (x − 4)(4 − 2x), find the x-coordinate for the relative maximum on the graph of f(x). The x coordinate of rel. max/min occurs when f ' (x) = 0 note this is a necessary condition, not a sufficient
Michele_Laino
  • Michele_Laino
as we can see from my drawing x=4 is a point of local maximum and x=2 is a point of local minimum
zeesbrat3
  • zeesbrat3
Right, now what if you wanted to use that graph to find the max of the original function f(x)?
Michele_Laino
  • Michele_Laino
the usage of my graph above is only to know that at x=4 there is a point of local maximum
Michele_Laino
  • Michele_Laino
and, of course at x=2 there is a point of local minimum
zeesbrat3
  • zeesbrat3
Okay, if we took a graph, like this one: How would you find the max and min of the original function?
Michele_Laino
  • Michele_Laino
it is simple, since from your graph we have three stationary points for f(x), furthermore only at x=1 there is a local maximum, since for point at left of x=1 the derivative is positive, so the function is incresing, whereas for point at right of x=1 the derivative is negative, so the function is decreasing. In other words there exists a neighborhood of x=1, such that the function is increasing at the left of x=1 and it is increasing at the right of x=1
Michele_Laino
  • Michele_Laino
increasing*
Michele_Laino
  • Michele_Laino
|dw:1437334409653:dw|
zeesbrat3
  • zeesbrat3
So 1 would be the max of the original function? Is there any way to find the min as well, or is there not enough information?
Michele_Laino
  • Michele_Laino
the minimum points are at x=0 and x=2, since for each of those points we have a neighborhood which contains point for which f is decreasing at the left, and points at which f is increasing at the right of the point which the neighborhood is referring to
zeesbrat3
  • zeesbrat3
Oh, thank you! I understand now. I wish I was able to give you more than one medal!
Michele_Laino
  • Michele_Laino
|dw:1437334708640:dw|
Michele_Laino
  • Michele_Laino
thanks! :)

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