Exponential Functions - Critical Points

[AustrianSaurkraut]

For the curve f(x) = lnx/x² with x>0. Find the coordinates of any critical points in exact answers only with no decimals.

For this question, I've been stuck on what to do after getting the derivative and how to solve for the critical points.
f'(x) = 2x(lnx)-1/x(x²)/(x²)²
f'(x) = 2x*lnx-x/x⁴
f'(x) = 2x*lnx-1/x³

Now from here is where I am clueless as to how to solve the equation, as I know you must set f'(x)=0 but I do not know how to solve with a lnx included. Any help will be greatly appreciated.

 

[lex]

[MATH]f'(x) = \frac{x^2 \cdot \left(\frac{1}{x}\right) -2x\ln{x}}{x^4}[/MATH]

 

[Dr.Peterson]

For the curve f(x) = lnx/x² with x>0. Find the coordinates of any critical points in exact answers only with no decimals.

For this question, I've been stuck on what to do after getting the derivative and how to solve for the critical points.
f'(x) = 2x(lnx)-1/x(x²)/(x²)²
f'(x) = 2x*lnx-x/x⁴
f'(x) = 2x*lnx-1/x³

Now from here is where I am clueless as to how to solve the equation, as I know you must set f'(x)=0 but I do not know how to solve with a lnx included. Any help will be greatly appreciated.

Please learn to use parentheses so that what you write is what you mean. You probably meant this:

For the curve f(x) = ln(x)/x² with x>0. Find the coordinates of any critical points in exact answers only with no decimals.

For this question, I've been stuck on what to do after getting the derivative and how to solve for the critical points.
f'(x) = (2x ln(x)-(1/x)(x²))/(x²)²
f'(x) = (2x*ln(x)-x)/x⁴
f'(x) = (2x*ln(x)-1)/x³

But that last line is clearly wrong, because you divided only one term of the numerator by x.

Of course, the first line is also wrong, as the terms in the numerator are in the wrong order.

But once you get the algebra right, you will have something that is easy to solve for x.

 

[AustrianSaurkraut]

Please learn to use parentheses so that what you write is what you mean. You probably meant this:

But that last line is clearly wrong, because you divided only one term of the numerator by x.

Of course, the first line is also wrong, as the terms in the numerator are in the wrong order.

But once you get the algebra right, you will have something that is easy to solve for x.

I just redid the question and solved for a proper X value. Thank you for pointing the the wrong order and the wrong cancelling, also sorry for the lack of parenthesis.

 

[topsquark]

As far as Calculus is concerned critical points are where the function value is equal to 0, does not exist, or where the derivative is 0, or where the derivative does not exist. (Sometimes add in the second derivative as well.) So you need to add x = 1 to your list of critical points.

If we add in coordinate geometry x and y intercepts are also critical points.

-Dan

 

[Steven G]

Dan, Why would think it is correct to include when the function value is 0 for a critical point? As you know, it is just the x-intercept.

 

[Dr.Peterson]

As far as Calculus is concerned critical points are where the function value is equal to 0, does not exist, or where the derivative is 0, or where the derivative does not exist. (Sometimes add in the second derivative as well.) So you need to add x = 1 to your list of critical points.

If we add in coordinate geometry x and y intercepts are also critical points.

-Dan

"Critical" is used in various ways, but in this context it has one meaning:

Calculus I - Critical Points

In this section we give the definition of critical points. Critical points will show up in most of the sections in this chapter, so it will be important to understand them and how to find them. We will work a number of examples illustrating how to find them for a wide variety of functions.

tutorial.math.lamar.edu

We say that x=c is a critical point of the function f(x) if f(c) exists and if either of the following are true.​

f′(c)=0 OR f′(c) doesn't exist ​

 

[topsquark]

Dan, Why would think it is correct to include when the function value is 0 for a critical point? As you know, it is just the x-intercept.

The collection of points I mentioned were those that were taught to me in school. In this case (0, 0) is the intercept but that's not true of most functions. These are all points to look at when sketching a graph so I thought I should add them to the list.

-Dan

 

[Dr.Peterson]

The collection of points I mentioned were those that were taught to me in school. In this case (0, 0) is the intercept but that's not true of most functions. These are all points to look at when sketching a graph so I thought I should add them to the list.

-Dan

I might call those "key points", or something like that.

And I have to admit, I have often wondered whether "critical point" was a fully defined technical term, or just another term commonly used in textbooks, that could be defined differently. Why aren't endpoints of the domain also included, as they too need to be checked when looking for extrema?

 

[nasi112]

I might call those "key points", or something like that.

And I have to admit, I have often wondered whether "critical point" was a fully defined technical term, or just another term commonly used in textbooks, that could be defined differently. Why aren't endpoints of the domain also included, as they too need to be checked when looking for extrema?

I had once argued with my professor that end points are critical points, but she said no. Then, we discovered that it all depends on the book definition. She taught us with a book that defines end points as critical points while she reads another book which does not include end points as critical points.

 

[Steven G]

I had once argued with my professor that end points are critical points, but she said no. Then, we discovered that it all depends on the book definition. She taught us with a book that defines end points as critical points while she reads another book which does not include end points as critical points.

That is a great story!

 

[Steven G]

I never heard end points being called critical points. However, if a function was defined on [a,b], then I told my students to find the absolute max or mins you need to check the critical points as well as the end points. I can see why someone would define end points as critical points. Then again, that allows +/- infinity to be c.p.s. Maybe call end points c.p.s on a closed/clopen interval?

 

[Dr.Peterson]

I never heard end points being called critical points. However, if a function was defined on [a,b], then I told my students to find the absolute max or mins you need to check the critical points as well as the end points. I can see why someone would define end points as critical points. Then again, that allows +/- infinity to be c.p.s. Maybe call end points c.p.s on a closed/clopen interval?

One answer that has been given is that an endpoint of the domain that is included in the domain fits the definition of a critical point because the function does not have a (two-sided) derivative at an endpoint. So technically there is no need to list endpoints separately, as they are critical points in that sense. (But this raises the question, Why shouldn't we say that a one-sided derivative is all you need at an endpoint to call it differentiable there, since nothing more is possible?)

But I don't know that I've ever seen a textbook express it that way, and it strikes me as a technicality that students would tend to miss if not told explicitly to check both critical points and endpoints, even if you count the latter as among the former.

Ultimately, the reason one would call an included endpoint a critical point is so that the term "critical point" would be identical to "point at which an extremum might be found", which is useful. I don't think I've found a textbook that does so, though; I'd like to see Nasi112's source.

The main point is that definitions are arbitrary; a word means whatever the general consensus is, and can't be proved by logic.