Homework Help: Some calculus exercises doubts

1. Aug 16, 2010

Taturana

1. The problem statement, all variables and given/known data

Problem 1
Find the equation of the tangent line on the following points of the curve.
$f(x) = \frac{1}{x - a}, a \in \mathbb{R} - \left \{ -2; 4 \} \right; x = -2, x = 4$

Problem 2
Find the equation of the line normal to the tangent line on the following points of the curve.
$f(x) = x^{2} - 1; x = 0$

2. Relevant equations

3. The attempt at a solution

Problem 1
[PLAIN]http://img21.imageshack.us/img21/4021/giflatex2.gif [Broken]

As you can see my solution is completely wrong ;(

Problem 2
[PLAIN]http://img46.imageshack.us/img46/4265/giflatex3.gif [Broken]

It will give a division by zero, what's going on here?

Last edited by a moderator: May 4, 2017
2. Aug 16, 2010

Staff: Mentor

Evaluate your function and its derivative, and then use the values in your formulas. For #2, since f'(x) = 2x, then f'(0) = 0. This says that the tangent line is horizontal when x = 0 (i.e., at (0, 0). What does this say about the normal to the curve at this point?

Last edited by a moderator: May 4, 2017
3. Aug 16, 2010

vela

Staff Emeritus
In problem 1, you simply forgot to substitute x=-2 in when you wrote in the expression for f'(-2) in the equation of the line.

In problem 2, you found the slope of the tangent to be given by f'(x)=2x. At x=0, you get f'(x)=0, which means the tangent line is horizontal. The normal, therefore, will be vertical. Vertical lines do not have a finite slope, so you can't use the point-slope formula for a line.

4. Aug 16, 2010

Staff: Mentor

For #1, evaluate f(-2) and f'(-2) before trying to find the equation of the tangent line. My answer agrees with the book's answer that you posted.

5. Aug 16, 2010

Taturana

Thank you for the answers, that helped me alot, now it works... =D

Just another question: how do I know when do I have to use the chain rule? There's any easy way of visualize it or is it just practice?

6. Aug 16, 2010

Staff: Mentor

Use the chain rule whenever you are differentiating a composite function. The chain rule actually comes into play in your first problem, but because the derivative of x - a is 1, the contribution from the chain rule isn't noticeable.

If you had f(x) = 1/(2x - a) = (2x - a)-1, then f'(x) = -1(2x - a)-2 * 2 = -2/(2x - a)2. The 2 that appears in the numerator of the last expression comes from the use of the chain rule, and is the derivative of 2x - a.

7. Aug 16, 2010

Taturana

Thank you but I have another doubt (I think its not necessary to open another thread).

suppose:

$$f(x) = x(3x-5)$$

I need to use the chain rule here, right?

$$f(x) = x u$$
$$u = 3x -5$$
$$\frac{df}{dx} = \frac{df}{du} \frac{du}{dx}$$
$$\frac{df}{du} = x$$
$$\frac{du}{dx} = 3$$
$$\frac{df}{dx} = 3x$$

I know that is not right, but why?

8. Aug 16, 2010

Staff: Mentor

You have another question, not a doubt. You have a doubt about something when you think you know what it is, but you aren't sure.

And yes, you should start a new thread when you have a new problem.
No. This is a product, not a composition, so you should use the product rule.
Here it is using the product rule.
f'(x) = x * d/dx(3x - 5) + (3x - 5) * d/dx(x)
= x * 3 + (3x - 5) * 1
= 3x + 3x -5 = 6x - 5.

As a check, f(x) = 3x2 - 5x, so f'(x) = 6x - 5.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook