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

- chestercat

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- zeesbrat3

- zeesbrat3

Please help

- 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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## More answers

- zeesbrat3

Because it wouldnt exist?

- Michele_Laino

since we can not divide by zero

- zeesbrat3

Sorry, I thought of limits when I said it doesn't exist since like you said, you can't divide by 0

- zeesbrat3

I think A i incorrect as well

- Michele_Laino

yes! I think so, since we can have this case:
|dw:1437332734543:dw|

- 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

So, since A and C are incorrect, the only valid answer would be B, yes?

- 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

oops.. f '(0)=0

- Michele_Laino

yes! option B is the correct one

- zeesbrat3

Thank you for clarifying C for me. :) Would you mind helping me with another?

- Michele_Laino

ok!

- zeesbrat3

Given f '(x) = (x − 4)(4 − 2x), find the x-coordinate for the relative maximum on the graph of f(x).

- 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

then I solve this inequality:
\[ - 2{x^2} + 12x - 16 \geqslant 0\]

- Michele_Laino

which is equivalent to this one:
\[{x^2} - 6x + 8 \leqslant 0\]

- zeesbrat3

I got \[-2(x-4)(x-2)\ge0\]

- Michele_Laino

yes! it is right!

- zeesbrat3

Then divide both sides by -2. So 4 is the max?

- zeesbrat3

Because it is the highest point

- Michele_Laino

we have to proceed with caution

- Michele_Laino

the solution of our inequlity is:
\[2 \leqslant x \leqslant 4\]

- Michele_Laino

inequality*

- Michele_Laino

so, we have thi drawing:
|dw:1437333653697:dw|

- Michele_Laino

where f '(x) is positive then f(x) is increasing
where f '(x ) is negative then f(x) is decreasing

- zeesbrat3

The number line is your checks to see whether it is min or max..

- Michele_Laino

yes!

- Michele_Laino

|dw:1437333827000:dw|

- 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

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

Right, now what if you wanted to use that graph to find the max of the original function f(x)?

- 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

and, of course at x=2 there is a point of local minimum

- zeesbrat3

Okay, if we took a graph, like this one: How would you find the max and min of the original function?

##### 1 Attachment

- 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

increasing*

- Michele_Laino

|dw:1437334409653:dw|

- 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

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

Oh, thank you! I understand now. I wish I was able to give you more than one medal!

- Michele_Laino

|dw:1437334708640:dw|

- Michele_Laino

thanks! :)

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