Question 1

What is the difference between critical numbers and critical points ?
Are they the same ?

Accepted Answer

Sometimes they're loosely used interchangeably, but typically when you're asked to identify a critical point, you're expected to provide the coordinates of the point (both x and y), and a critical number is just the x value: the place on the number line where f' is zero or undefined.

Question 2

How do we know which point is going to be the maximum or minimum point?

Accepted Answer

Assuming you have figured out what the critical points are, you can just take any one convenient number between each two neighbouring critical points and *evaluate the derivative function f'(x*) at those points that you have chosen.

Then you look at every critical point and check—using your new data—if the derivative is negative before it but turns positive after it (makes it a *minimum* point) _or_ is positive before but turns negative (*maximum*) _or_ doesn't change sign, in which case you don't care about that critical point.

[*Justification*] That works, because where the derivative is positive, the function is increasing and where it is negative, the function is decreasing. If it steadily increases up to some point and then starts dropping down, that's obviously a maximum. The opposite case corresponds to a minimum.

Question 3

at the juncture where we analyze which of the terms e^(-2x^2)* (1-4(x^2)) is 0, it is mentioned that the 'e' term can never be 0 but e^(-∞) is 0 , why is this possibility neglected ,especially when the range of the function is not mentioned

Accepted Answer

−∞ is not a number, so you will never have that in the equation. You can approach it (as in the case of a limit)  but never reach −∞. Thus, you won't have a critical point where x=±∞ because there is no such point (it is only approached, but never reached).

Question 4

At 2:50, shouldn't d/dx e^-2x^2 = -2e^-2x^2*(-4x)?

Accepted Answer

I put it into wolfram alpha and got: Possible derivation:
d/dx(x/e^(2 x^2))
Rewrite the expression: x/e^(2 x^2) = x/e^(2 x^2):
  =  d/dx(x/e^(2 x^2))
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = e^(-2 x^2) and v = x:
  =  x (d/dx(e^(-2 x^2)))+e^(-2 x^2) (d/dx(x))
Using the chain rule, d/dx(e^(-2 x^2)) = ( de^u)/( du) ( du)/( dx), where u = -2 x^2 and ( d)/( du)(e^u) = e^u:
  =  (d/dx(x))/e^(2 x^2)+(d/dx(-2 x^2))/e^(2 x^2) x
Factor out constants:
  =  (d/dx(x))/e^(2 x^2)+(-2 d/dx(x^2) x)/e^(2 x^2)
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x:
  =  (d/dx(x))/e^(2 x^2)-(2 x 2 x)/e^(2 x^2)
Simplify the expression:
  =  -(4 x^2)/e^(2 x^2)+(d/dx(x))/e^(2 x^2)
The derivative of x is 1:
  =  -(4 x^2)/e^(2 x^2)+1/e^(2 x^2)
Simplify the expression:
Answer: |  
 |   =  e^(-2 x^2)-(4 x^2)/e^(2 x^2)

Question 5

How can I find critical values of a function when the zeros of the first derivative of that function are imaginary numbers? For example, the function is 7x^3 - 3x^2 + x - 15.

Accepted Answer

There is no rule saying a function has to have ANY critical points. For example, 
f(x)= 3x+7 has no critical points.

Your particular function has no REAL critical points. Most likely, that is what an introductory calculus course would be asking about, so you would most likely be expected to say it had no real critical points.
 
However, it is completely valid to have nonreal critical points. All you do is find the nonreal zeros of the first derivative as you would any other function. You then plug those nonreal x values into the original equation to find the y coordinate.
 
So, the critical points of your function would be stated as something like this:
There are no real critical points.
There are two nonreal critical points at:
x = (1/21) (3 -2i√3), y= (2/441) (-3285 -8i√3)
and
x = (1/21) (3 + 2i√3), y= (2/441) (-3285 + 8i√3)

Question 6

at  4:28 why the e^-2x^2 cant be zero ?

Accepted Answer

The exponential function (when considered a function of real numbers) is always positive. That is, for every real number u, we have exp(u) > 0. Now let u = -2x², which is a real number whenever x is a real number. It follows that exp(u) = exp(-2x²) > 0. Hence never zero.

The fact that exp(u) > 0 for every real number u requires a proof in itself, but I will not provide one here (unless you really want me to).

The fact that `exp(u) > 0` for every real number `u` requires a proof in itself, but I will not provide one here (unless you really want me to).

Question 7

If you wanted to find a critical POINT (not number) would you just take the 'x' value -the critical number, and plug it into the original equation to get a 'y' value?

Accepted Answer

Yes, that is how you would find a critical point. You get the x from the first derivative and the y from the original function.

Question 8

Doesn't f(c) need to exist for x=c to be a critical point of the function?  This example doesn't highlight this part of the critical point definition.

Accepted Answer

Hi there,
When defining a critical point at x = c, c must be in the domain of f(x). So therefore, when you are determining where f'(c) = 0 or doesn't exist, you aren't included discontinuities as possible critical points.  Here is an example.

f(x) = x^(2/3). The domain here is all real numbers.
f'(x) = 2/3 x^(-1/3) or 2/(3x^1/3) . (or 2 divided by 3 cube root x)

x = 0 is not in the domain of the derivative, but it is in the domain of the function.  Therefore, x = 0 is considered a critical point of f(x).

Hope this helps

AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB > Unit 5

Finding critical points

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