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## Calculus 1

### Course: Calculus 1>Unit 5

Lesson 2: Extreme value theorem and critical points

# Finding critical points

Sal finds the critical points of f(x)=xe^(-2x²). Created by Sal Khan.

## Want to join the conversation?

• What is the difference between critical numbers and critical points ?
Are they the same ? •   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.
• How do we know which point is going to be the maximum or minimum point? •  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.
• Are all global maximums and minimums also local? • 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 • −∞ 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).
• 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. • 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)
• At , shouldn't d/dx e^-2x^2 = -2e^-2x^2*(-4x)? • 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:
| = e^(-2 x^2)-(4 x^2)/e^(2 x^2)
• At Sal mentioned that 'c' is a critical number 'iff' f'(c) = 0 or f'(c) undefined.

But I did few exercises on Khan Academy that ignores points when f'(x) is undefined and states that answer as incorrect.

*For example: *
When I'm doing a quiz on Khan Academy I see this:
h(x) = e^2x/(x-3) and a question asking to find critical points.
If you take derivative of that you would get:
h'(x) = e^2x(2x - 7)/(x - 3)^2

Now clearly h'(x) is 0_ when x = 3.5, but also _h'(x) is undefined when x = 3

Yet if you present that both x=3.5 and x=3 are critical number quiz returning answer than 3 is an incorrect answer.
It's not just isolated quiz, I did a bunch of them and every time you try to present critical x when f'(x) is undefined it renders my answer incorrect. What am I missing?

P.S.: I got it a minute after I submitted the question. Apparently x is not a critical point in this case, because when x=3 not only it's derivative h'(3) is undefined but also main function h(3) is also undefined. So it is not considered as a critical point in this case. • at why the e^-2x^2 cant be zero ? • 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).
• 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? • 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. • 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