Basic Function Problem With a Given Derivative

[DMOC]

Homework Statement

Let h be a function defined for all x =/= such that h(4) = -3 and the derivative of h is given by h ' (x) = (x$$^{2}$$-2)/x for all x=/=0.

a) Find all values of x for which the graph of h has a horizontal tangent, and determine whether h has a local maximum, a local minimum, or neither at each of these values. Justify your answers.

b) On what intervals, if any, is the graph of h concave up? Justify your answer.

c) Write an equation for the line tangent to the graph of h at x = 4.

d) Does the line tangent to the graph of h at x = 4 lie abov eor below the graph of h for x>4? Why?

Homework Equations

I don't think there are any.

The Attempt at a Solution

Let's try part a, and I'll do the other three once I get part a down.

So for part a, I need to find the values of x for which the graph of h has a horizontal tangent. Unfortunately, I'm not sure how to accomplish this, and I know I can't use a calculator.

So I'm thinking that the correct way to do this problem is to find the original graph of h(x). I have h'(x). However, I'm not sure how to actually find the original equation. When I do some fiddling around with h'(x), I get this:

h'(x) = x - 2x$$^{-1}$$

And I'm not sure how to find the original equation, or even if that's the correct way to do this problem.

 

[LCKurtz]

Hint: What is the value of the derivative of a function at a point where it has a horizontal tangent line?

Second hint: What does the second derivative tell you about concavity?

 

[DMOC]

The value of the derivative of a function at a point where it has a tangent line is zero.

If the second derivative is positive, it's concave UP, or like a smiley face. Actually, that's how I remember second derivatives. I just draw two pluses and a smile, and that's concavity up with a positive second derivative. A negative second derivative is a frown (concavity down) with two negative signs as "eyes."

So I should try to set the first derivative to equal zero, methinks.

The first derivative is zero when x = $$\sqrt{2}$$

Am I on the right track? Thanks incidentally.

 

[LCKurtz]

Right track, yes. Does x² = 2 have only that one solution?

 

[DMOC]

Oops, I forgot to put in a plus or minus (+/-). Is that what you were thinking of?

 

[LCKurtz]

Yes.

 

[DMOC]

So the answer to part a.

(a) Find all values of x for which the graph of h has a horizontal tangent, and determine whether h has a local maximum, a local minimum, or neither at each of these values. Justify your answers.

Answer: $$\sqrt{2}$$ and -$$\sqrt{2}$$ are the two values of x for which the graph of h has a horizontal tangent.

Furthermore, because $$\sqrt{2}$$ happens to be positive, then the second derivative will also be positive. Therefore, at the point where x=$$\sqrt{2}$$, h will have a local minimum, because a positive first derivative means a concavity up where there is a minimum.

When x=-$$\sqrt{2}$$, there is a concavity downwards, so there will be a maximum at that point.

How does this look?

 

[LCKurtz]

Better check your second derivative and signs.

 

[DMOC]

Uh oh ... is it something like the other way around? Or is this a completely different issue?

 

[LCKurtz]

You have the theory right. Just check your second derivative is correct and that your concavity at the two points is correct. Right now it isn't.

 

[DMOC]

I took the second derivative of h.

h"(x) = (x(2x-1) - (x$$^{2}$$-x))/x$$^{2}$$

Which simplifies to ... 1.

I'm getting a bit lost here.

 

[LCKurtz]

Your formula for h'' isn't right. Take the derivative more carefully. Or better yet, write
h'(x) = x - 2x^-1 before you differentiate it.

 

[DMOC]

I didn't think h" was right so I solved it correctly the second time I did it (before reading your post).

I was able to solve this problem completely though.

For part b, I used the second derivative to equal zero.

For part c, I used the point slope formula (y - y1) = m(x - x1).

Part d, answer is below since the graph is concave upwards so if it's concave up, the tangent would, of course, be below the concave.

Thanks for your help, you got me started on the right track.