I'm trying to teach myself some calculus for fun and could use some
insights.  If you could help me with a couple of things I'm having
trouble with I'd really appreciate it.

Suppose the cost function (in$) to manufacture x items is given by:

C(x) = ((1/3)x^3) ? ((2x)^2) + kx + 10.

1.	How do I find the value of k so that C(x) has exactly one
horizontal tangent line?

2.	How do I identify the value of x at the place where the above
unique horizontal tangent line occurs?

3.	How then do I compute the cost in manufacturing the item, if x is
the value you computed in question 2 and k is the value computed in
question 1.

4.	How do I find the value of k so that y = 2x is a tangent line to
the parabolic curve y = x^2 + k.

5.	A manufacturer of widgets that the cost to produce x widgets can be
calculated by:  C(x) = ((.1x)^2)-10x+10000.  What is the marginal
cost?

6.	Still working on 5, what is the marginal cost when x is 10?

7.	Still working on 5, how do I find the value for x for which the
cost is the maximum?

8.	If y = x^x then how do I find (dy/dx) at x = e?

9.	How about if y = Ln[Ln[x]] then how do I find (dy/dx) at x = e?

Thanks!

Hi alexinia!
Here are the answers to your questions.

Question 1
I think that in this equation you have a typo which is an extra
parenthesis in the second term. That is, you wrote (2x)^2 (which would
be 4x^2) when you intended to write 2x^2. In any case, if the term was
correct with 4x^2, the procedure to solve this problem is the same.
So, I assume the equation is:

C(x) = (1/3)x^3 ? 2x^2 + kx + 10

A tangent line to this curve would have the same slope as this curve
at the point it touches it. Therefore, in order to answer this
question, we must find the first derivative of C(x) (which is its
slope) and see where this derivative is equal to zero (recall that
when the slope in a line is zero, the line is horizontal). So we have:

C'(x) = x^2 - 4x + k

Equating this to zero gives:

x^2 - 4x + k = 0

Notice that this is a quadratic equation, which has at most two
solutions. That it is, there are at most two points at which the slope
of this curve is zero and thus the tangent line is horizontal. We want
to find the value of k such that there is only ONE solution to this
equation; that is, there is only one point at which the tangent line
to that point is horizontal. Using the rule to solve quadratic
equations we get that the two solutions are:

x1 = ( 4 + sqrt(16-4k) )/2
x2 = ( 4 - sqrt(16-4k) )/2

[sqrt() stands for square root of]
So, these two solutions clearly become the same one (making the
solution unique) when sqrt(16-4k) = 0; otherwise they are different
and thus you'll have two different places at which the slope is zero.
So we solve:

sqrt(16-4k) = 0
16-4k = 0
k = 4

When k=4, x1=x2=2. Therefore, when k=4, there is only one point at
which the original curve has an horizontal tangent line.


Question 2

As we found above, the slope is zero when x = 2. So this is the value
of x where the horizontal line occurs.


Question 3

In order to answer this, we simply plug k = 4 and x = 2 in the
original equation C(x)

 (1/3)x^3 ? 2x^2 + kx + 10
=(1/3)2^3 ? 2*2^2 + 4*2 + 10
=8/3 - 8 + 8 + 10
=38/3

That's the cost of manufacturing the item.


Question 4

We know here that the slope of the tangent line is 2. Therefore, we
must first find at what point in the parabolic curve this line is
tangent. That is, we first find where the slope of the parabolic curve
is also equal to 2. The curve is:

y = x^2 + k

The slope is the first derivative:

y' = 2x

Therefore, the slope of the curve is equal to 2 when x=1. Plugging x=1
into the parabolic curve gives

y = 1 + k

Therefore, the tangent line passes through the point (1, 1+k). Now,
plugging x=1 into the tangent line, which is y=2x, we get that y=2.
Therefore, the tangent line passes through the point (1, 2). Since
this point must be the same as the previous one (because x=1 in both
cases), we conclude that:

2 = 1 + k
k = 1

So y=2x is a tangent line to the parabolic curve y=x^2 + 1


Question 5

I think there is some kind of typo in the equation you posted here,
because the solutions give negative marginal costs (for question 6),
and that there is no maximum cost (question 7). Please re-check that
you've written it correctly.

Anyway, I'll solve it here as if the equation was correct so that you
have an idea of the procedure that must be followed to solve this
problem. You have:

C(x) = (.1x)^2-10x+10000

which is the same as:

C(x) = 0.01x^2 - 10x + 10000

Given a cost function, the marginal cost is the first derivative of
the cost function. Therefore, the marginal cost is:

C'(x) = 0.02x - 10


Question 6

This question simply requires you to plug x=10 into the marginal cost
function. This gives:

C'(10) = 0.02*10 - 10
       = 0.2 - 10
       = -9.8

(Negative marginal costs make no sense, that's why I think there is a
typo i nthe equation.


Question 7

If the equation you wrote is correct, the cost function is a convex
parabola. These have no maximum - they go to infinity as x increases.


Question 8

In this question and the following one, the commenter shockandwave-ga
gave the correct answer. Using the property that exp and ln cancel
each other out, so that exp(ln(x))=x, we can rewrite the this equation
as:

x^x = exp(ln(x^x))

Then, we use the logarithm property ln(x^y)=y*ln(x) to rewrite the
above equation as:

exp(ln(x^x)) = exp(x*ln(x))

We can find the derivative of the function exp(x*ln(x)) using the chain rule.

Chain Rule Tutorial
http://www.1728.com/chainrul.htm

Setting u=x*ln(x), we get that the derivative must be:

[exp(u)]'*u'

Since the derivative of exp(x) is also exp(x), then [exp(u)]' =
exp(u). Now we must find u', which is the derivative x*ln(x). Here we
use the multiplication rule:

u' = [x*ln(x)]' = 1*ln(x) + x*(1/x)
                = ln(x) + 1

So now we can find the answer:

[exp(u)]'*u' = exp(u)*u' = exp(x*ln(x))*(ln(x)+1) = (x^x)(ln(x)+1)

Evaluating x=e, and using the fact that ln(e)=1, we get:

(e^e)(1+1) = 2e^e

So 2e^e is the derivative of y=x^x at x=e.


Question 9

This question can be solved again using the chain rule. We have

y = ln(ln(x))

Setting u=ln(x), we have that:

y = ln(u)

So, then:

y' = [ln(u)]' * u'

Since u=ln(x), then u'=1/x. Also, [ln(u)]'=1/u. Therefore, we have:

y' = (1/u) * (1/x) = (1/ln(x)) * (1/x)

When we set x=e, we get that:

y' = (1/1) * (1/e) = 1/e

SO the derivative of y at x=e is 1/e.


I hope this helps! If you have any questions regarding my answer,
please don't hesitate to request a clarification. Otherwise I await
your rating and final comments.

Best wishes!
elmarto

This is better than getting a tutor because I have something to look
back on later -- thank you so much for your help!

I'll do number 8 cause it looks fun.

Rewrite x^x as follows, and (this is a good idea for any a^x type
derivatives or integrals) x^x = exp(ln(x^x)). Notice that exp and ln
"undo" each other so I clearly made no change. Then use the property
of logarithms that says ln(x^x)= x*ln(x). So we have x^x =
exp(x*ln(x)). Remembering the derivative of exp(U)=exp(U)*U'.  In this
case U = x*ln(x) and U'= x*(1/x)+ln(x)*1 (using the product rule for
derivatives.) So derivative of x^x = exp(x*ln(x))*(1+ln(x)) = x^x *
(1+ln(x)). Evaluated at x=e, leads to: 2*e^e.

Ok, i'll do number 9 too. 

remember the derivative of ln(U) = 1/U * U'. In this case U=ln(x) and U' = 1/x.
So derivative of ln(ln(x)) = 1/(lnx) * (1/x). Evaluated at x=e, leads to, 1/e.

Okay, okay, I'll do number one as well. But these better not be homework!

There are numerous ways to reason your way through this problem.
Here's how I approached it.

The slope of the tangent line is given by the derivative C'(x).
Horizontal tangent line means derivative is equal to zero. So.. Taking
the derivative and setting it to zero, C'(x) = x^2 - 4x + k = 0. Take
a moment to consider what we are talking about. Its a family of upward
opening parabolas. Consider, if the vertex is above the x-axis, it
won't have any real roots - and hence won't have any horizontal
tangents. If the vertex is below the x-axis, then it will have two
real roots and hence two horizontal tangents. Our task is to then
choose k such that its vertex falls right onto the x-axis.  Setting
C''(x) = 0 and solving for x will give us the x value corresponding to
the vertex. C''(x) = 2x - 4 = 0 implies x=2 is the vertex. So that
parabola C'(x) needs to contain the point (2,0). C'(2)=0 Leads to
(2)^2 - 4(2) + k = 0 ...same as... -4+k=0 and hence k=4.