# Simple question, tangent line

1. Aug 22, 2006

### DB

really isnt calculus but thats the name of my class so, anyway im having a problem here. i think its pretty simple im just missing something. im supposed to find the slope of the tangent line using one point and:

slope of tangent line =

$$\frac{f(a+h)-f(a)}{h}$$

$$h\rightarrow0$$

find slope of tangent line:

$$f(x)=\frac{1}{x}$$

at a=2

so,

$$\frac{1}{2+h}-\frac{1}{2}*\frac{1}{h}$$

common denominator X by (1+h)

$$\frac{1}{2+h}-\frac{1+h}{2+h}*\frac{1}{h}$$

$$\frac{1-1-h}{2+h}*\frac{1}{h}$$

$$\frac{-2-h}{2+h}*\frac{1}{h}$$

$$\frac{-1(2+h)}{2+h}*\frac{1}{h}$$

$$\frac{-1}{h}$$

make "h" zero and im stuck.....

2. Aug 22, 2006

### d_leet

Ummm... Is 1/2 the same as (1+h)/(2+h)?

3. Aug 22, 2006

### DB

thanks

okay stupid mistake, now im stuck again tho

$$\frac{2}{2(2+h)}-\frac{(2+h)}{2(2+h)}*\frac{1}{h}$$

comes out to

$$\frac{-1}{h}}*\frac{1}{h}$$

gives me -1/h which doesnt work, darn

4. Aug 22, 2006

### d_leet

You need a factor of h in the first denominator for the denominators to be equal.

5. Aug 22, 2006

### DB

i dont understand

6. Aug 22, 2006

### d_leet

The denominator of the first term expanded is (4+2h) while the second term's denominator is (4h+2h2)

7. Aug 22, 2006

### DB

but those are different terms, arent i looking for a common denominator?

8. Aug 22, 2006

### benorin

If $$f(x)=\frac{1}{x},$$ then

$$\frac{f(a+h)-f(a)}{h} = \frac{\frac{1}{2+h}-\frac{1}{2}}{h} = \frac{1}{h(2+h)}-\frac{1}{2h} = \frac{2-(2+h)}{2h(2+h)}$$
$$=-\frac{1}{2(2+h)}\rightarrow -\frac{1}{4} \mbox{ as }h\rightarrow 0$$

9. Aug 22, 2006

### d_leet

Yes which is why you need to multiply the first term by h/h so that you have a common denominator.

10. Aug 22, 2006

### DB

thanks for the replies, but i dont understand how the c.d is 2h(2+h)

11. Aug 24, 2006

### DB

finally, got it

12. Aug 24, 2006

### Skhandelwal

I was wondering one thing about tangent line. Since tangent lines can also be secant line, I actually had a question about secant line. Is it possible for a secant line to intersect 2 consecutive points of a curve? I guess what I am trying to get to is, in a curve, can two, next to each other points be in a straight line?

13. Aug 24, 2006

### d_leet

What do you mean by two consecutive points? Between any two points on a curve there are an infinite number of other points so there really isn't such a thing as consecutive points in the way you are probably thinking. But if you mean like 2 points on a curve where say x = 1 and x = 2, then sure you just have to find a function with equal derivatives at those points, I can't think this through completely right now but that's the gist of it.

14. Aug 24, 2006

### Skhandelwal

Can you give me an example?

15. Aug 24, 2006

### d_leet

Can you give me one of what you mean by consecutive points?

16. Aug 25, 2006

### bomba923

Are you referring to something like y=sin(2πx) ?
It has equal derivatives at x=1 and x=2 (i.e., y'(1)=y'(2)=2π)

17. Aug 25, 2006

### d_leet

No not quite, since that gives a nice counterexample to what I was thinking, errr... I guess not what i was thinking but what I wrote since what I was thinking was that if there are two points on a curve, if the derivative at both of these points are equal and the tangent lines at each point have the same y intercept then they must be the same line and the tangent line would be a secant line. With a sine function like the one you mention you'll have the equal derivatives and hence slopes of the tangent lines are the same but the y intercepts are different. But again with that same function consider y=1 this will be tangent to that curve for x=1/4+n where n is an integer, but this probably isn't quite what the original poster was talking about and isn't exactly what i was either. If you still don't get what I mean I'll try and write up a better post and find a good example of this tomorrow.

18. Aug 25, 2006

### bomba923

Perhaps you were thinking of y=cos(2πx) ?
Where y'(1)=y'(2)=0 and the single line y=1 is tangent to y(x) at x=1 and x=2
(generally, as y=1 is tangent to y(x) here for all integer 'x')

Last edited: Aug 25, 2006
19. Aug 25, 2006

### d_leet

Well y=1 is tangent to sin(2πx) for all x of the form 1/4+n where n is an integer, isn't it? Because you would have sinn(π/2 + 2πn) which is 1 for every integer n, isn't it? You're example works to, and is better than mine since you don't have to have the 1/4 part.