Find all values of x for which the tangent line to y = f(x) is horizontal

[getoutofmylaboratory]

Find all values of x for which the tangent line to y = f(x) is horizontal.

(a) f(x) = x^2e^-2x
(b) f(x) = 2 ln x^3

I'm really not sure what to do here. The slope will be 0 since the line is horizontal. Since this chapter is about differentiation, I'm assuming I would have to come up with f prime of both, although, not really sure how to do that.

 

[getoutofmylaboratory]

I'm not exactly getting your confusion.

Are you able to find f'(x) in both cases?

To put it simply, I've somehow landed in a Calc I class without pre-calc or trig. I'm playing a lot of catch up. To answer your question, I'm not sure how to come up with f' for either.

From what little I understand:

d/(dx)(e^(-2x)x^2) = x^2[d/(dx)(e^-2x)] + e^-2x[d/(dx)(x^2)]

 

[Deleted member 4993]

To put it simply, I've somehow landed in a Calc I class without pre-calc or trig. I'm playing a lot of catch up. To answer your question, I'm not sure how to come up with f' for either.

From what little I understand:

d/(dx)(e^(-2x)x^2) = x^2[d/(dx)(e^-2x)] + e^-2x[d/(dx)(x^2)]

Why did you accept that?

This is not catch-up - this is swimming up-stream!

But it is your grade - and your learning - after all.

Back to the problem:

You have applied product rule - correctly.

Now calculate:

\(\displaystyle \frac{d}{dx} x^2 \ = \ ??\) ............. need to use \(\displaystyle \frac{d}{dx} x^n \ = \ n * x^{(n-1)}\)

and

\(\displaystyle \frac{d}{dx} e^{-2*x} \ = \ ??\) ..........................need to use chain rule and \(\displaystyle \frac{d}{dx} e^{x} \ = \ e^x\)

 

[getoutofmylaboratory]

Why did you accept that?

My degree calls for 6 credits of College Algebra and above. At the time, I didn't see any courses available that met that requirement other than Calculus I.

This is not catch-up - this is swimming up-stream!

Yes, I agree. It's definitely not the ideal way I'd have liked to do it.

But it is your grade - and your learning - after all.

Fortunately, my grade isn't suffering...at least not yet. Mid-term exam coming up in about a week, so we'll see how that goes.

Back to the problem:

You have applied product rule - correctly.

Now calculate:

\(\displaystyle \frac{d}{dx} x^2 \ = \ ??\) ............. need to use \(\displaystyle \frac{d}{dx} x^n \ = \ n * x^{(n-1)}\)

and

\(\displaystyle \frac{d}{dx} e^{-2*x} \ = \ ??\) ..........................need to use chain rule and \(\displaystyle \frac{d}{dx} e^{x} \ = \ e^x\)

So, I did get some help from a coworker, which got me to:

f(x) = (x^2)[e^(-2x)]
f'(x) = 2x[e^(-2x)]

0 = 2x[e^(-2x)]
x = 0

Does that seem right?

that's a hard way to go. You're going to need that trig so I would get on it. You're going to need to be pretty facile with quadratics too so I'd get on those as well.

Tell me about it Yeah, it didn't take me long to figure out that trig was going be a key player in this course. If you know of any resources that might be helpful, do let me know. For starters, I suppose it would help to have a sort of cheat sheet to help me memorize common trig conversions. I'm aware of SOHCAHTOA.

What you posted above for the derivative is correct. So take it to the next step and take those derivatives

d/dx (e^ax) = a e^ax

I'm going to assume you can differentiate a polynomial term.

d/dx (ln(x)) = 1/x

it looks like you might need the chain rule too

d/dx(g(f(x)) = dg/dx(f(x)) df/dx(x)

yell back if this makes no sense

yelling back!

My coworker was showing me part b of that question, too, but I am probably missing something here:

2x ln e^-2x = ln 0
-4 x^2 = -1
x^2 = -1/4

x +/- 1/2
x = 0

I'm really lost on it, though.


Side note, are you all using a different editor to output the equations correctly? I don't see any options for that.

 

[JeffM]

With respect to trig functions, there are a number of handy summaries on the web, but this will get you started
http://en.wikipedia.org/wiki/Trigonometric_functions

However, you probably should do some practice on converting trig expressions from one form to another.

This thread is getting confusing so I am not sure I am on the right page.

\(\displaystyle y = x^2e^{-2x}.\)

A lot of the mechanics of differential calculus is memorizing some basic derivatives, doing substitutions, and using the chain rule.

Here goes.

\(\displaystyle y = uv \implies \dfrac{dy}{dx} = u * \dfrac{dv}{dx} + v * \dfrac{du}{dx}.\) One of the basic derivatives to memorize.

So, for this problem which involves a product, that suggests the substitutions \(\displaystyle u = x^2\ and\ v = e^{-2x}.\)

\(\displaystyle r = s^n \implies \dfrac{dr}{ds} = ns^{(n - 1)}.\) Another basic derivative to memorize. So \(\displaystyle u = x^2 \implies \dfrac{du}{dx} = 2x.\)

Now here is another basic derivative to memorize: \(\displaystyle p = e^q \implies \dfrac{dp}{dq} = e^q.\)

But we don't have something quite that simple. So we need another substitution.

\(\displaystyle w = -2x\ and\ v = e^{-2x} \implies v = e^w \implies \dfrac{dv}{dw} = e^w.\)

But we need dv/dx, not dv/dw. Now we use the chain rule, which is a key idea in calculus.

\(\displaystyle \dfrac{dv}{dx} = \dfrac{dv}{dw} * \dfrac{dw}{dx}.\) So what is dw/dx?

\(\displaystyle w = -2x \implies \dfrac{dw}{dx} = - 2.\)

OK Now we have built all the pieces needed

\(\displaystyle \dfrac{dy}{dx} = u * \dfrac{dv}{dx} + v * \dfrac{du}{dx} = u * \dfrac{dv}{dw} * \dfrac{dw}{dx} + v * \dfrac{du}{dx} = x^2 * e^w * (-2) + e^{-2x} * 2x = 2x\left(-xe^w + e^{-2x}\right).\) Do you see where all the pieces came from?

However the above is not quite right because we do not have the derivative completely in terms of x; there is that w in there. So substitute back.

\(\displaystyle w = - 2x\ and\ \dfrac{dy}{dx} = 2x\left(-xe^w + e^{-2x}\right) \implies \dfrac{dy}{dx} = 2x\left(-xe^{-2x} + 2e^{-2x}\right) = 2xe^{-2x}(1 - x).\)

Now proceed.

I hope doing this step by step helps you to see why I say memorization of basic derivatives, substitution, and the chain rule are all essential in calculus.

Edit: Studying calculus is hard enough. Don't worry about the formatting editor, which is fussy.

 

[getoutofmylaboratory]

Thanks for that informative post, JeffM. Give me a while to digest everything there