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zeesbrat3
 one year ago
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].
zeesbrat3
 one year ago
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].

This Question is Closed

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 @mathstudent55 @kropot72 @Kash_TheSmartGuy

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2hint: 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

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0Because it wouldnt exist?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2since we can not divide by zero

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, I thought of limits when I said it doesn't exist since like you said, you can't divide by 0

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0I think A i incorrect as well

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! I think so, since we can have this case: dw:1437332734543:dw

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0So, since A and C are incorrect, the only valid answer would be B, yes?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2another 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
 one year ago
Best ResponseYou've already chosen the best response.2yes! option B is the correct one

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for clarifying C for me. :) Would you mind helping me with another?

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0Given f '(x) = (x − 4)(4 − 2x), find the xcoordinate for the relative maximum on the graph of f(x).

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2In 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
 one year ago
Best ResponseYou've already chosen the best response.2then I solve this inequality: \[  2{x^2} + 12x  16 \geqslant 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2which is equivalent to this one: \[{x^2}  6x + 8 \leqslant 0\]

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0I got \[2(x4)(x2)\ge0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! it is right!

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0Then divide both sides by 2. So 4 is the max?

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0Because it is the highest point

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we have to proceed with caution

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the solution of our inequlity is: \[2 \leqslant x \leqslant 4\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so, we have thi drawing: dw:1437333653697:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2where f '(x) is positive then f(x) is increasing where f '(x ) is negative then f(x) is decreasing

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0The number line is your checks to see whether it is min or max..

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437333827000:dw

perl
 one year ago
Best ResponseYou've already chosen the best response.1Given f '(x) = (x − 4)(4 − 2x), find the xcoordinate 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
 one year ago
Best ResponseYou've already chosen the best response.2as we can see from my drawing x=4 is a point of local maximum and x=2 is a point of local minimum

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0Right, now what if you wanted to use that graph to find the max of the original function f(x)?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the usage of my graph above is only to know that at x=4 there is a point of local maximum

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2and, of course at x=2 there is a point of local minimum

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0Okay, if we took a graph, like this one: How would you find the max and min of the original function?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2it 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437334409653:dw

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.2the 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh, thank you! I understand now. I wish I was able to give you more than one medal!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437334708640:dw
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