# Tangent of a Curve

• Oct 11th 2007, 06:27 PM
Truthbetold
Tangent of a Curve
Determine whether he curve ha a tangent at the indicated point. If it does, give its slope. If not, explain why not?
f(x)=\left\{\begin{array}{rr}x&\ 2-2x-x^2 {if }x\ge0\\ -x&\2x-2{if } x > 0\end{array}\right. there is the messed up LaTex if anyone could/will/may :) etc, answer this. thank you!
The above is at x= 0. The second inequality is supposed to have an > or = sign.
Ah.

I couldn't get it. :( The LaTEx, I mean.
Here it is in math:
f(x)= 2-2x-x^2, x < 0
piecewise 2x+ 2, x > or = 0.
Graphola!
http://www.geocities.com/onlychristislife/graph56.jpg

I circled the point which I think they are talking about for the tangent.
If there isn't one, figuring out the slope will be messy. However its done.

Thanks!
• Oct 11th 2007, 06:38 PM
Jhevon
Quote:

Originally Posted by Truthbetold
Determine whether he curve ha a tangent at the indicated point. If it does, give its slope. If not, explain why not?
f(x)=\left\{\begin{array}{rr}x&\ 2-2x-x^2 {if }x\ge0\\ -x&\2x-2{if } x > 0\end{array}\right. there is the messed up LaTex if anyone could/will/may :) etc, answer this. thank you!
The above is at x= 0. The second inequality is supposed to have an > or = sign.
Ah.

I couldn't get it. :( The LaTEx, I mean.
Here it is in math:
f(x)= 2-2x-x^2, x < 0
piecewise 2x+ 2, x > or = 0.
Graphola!
http://www.geocities.com/onlychristislife/graph56.jpg

I circled the point which I think they are talking about for the tangent.
If there isn't one, figuring out the slope will be messy. However its done.

Thanks!

we have a cusp (sharp turn) at the indicated point. a function is not differentiable at its cusp. well, i think this is a cusp, it's definitely a sharp turn though

you can show that the derivative limit does not exist for that point
• Oct 11th 2007, 06:48 PM
Truthbetold
So no tangent at sharp turn, got it.
Quote:

you can show that the derivative limit does not exist for that point
How do you do that?

Since I have to find the slope, how do I find the slope?

Edit to the graph, the second blue dot goes to -33, not -100.

Slope formula is
lim (f(a + h) - f(a))/h
h-> 0

I believe this to be for the poitns (0, 2) and (5,-35), but I don't know.
The teacher didn't have much time to teach today.
• Oct 11th 2007, 07:17 PM
Jhevon
Quote:

Originally Posted by Truthbetold
So no tangent at sharp turn, got it.
How do you do that?

Since I have to find the slope, how do I find the slope?

Edit to the graph, the second blue dot goes to -33, not -100.

Slope formula is
lim (f(a + h) - f(a))/h
h-> 0

I believe this to be for the poitns (0, 2) and (5,-35), but I don't know.
The teacher didn't have much time to teach today.

it is better to use this formula here, i think. it is an equivalent formula.

$\displaystyle f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}$

now, we are concerned with what happens when x approaches 0. but we have a piece wise function such that x approaches 0 in different ways form the left and the right. thus we must find the left and right hand limits of the derivative limit.

$\displaystyle \lim_{x \to 0^+} \frac {f(x) - f(0)}{x - 0}$ ........we are approaching from the right, here $\displaystyle f(x) = -x^2 - 2x + 2$

and

$\displaystyle \lim_{x \to 0^-} \frac {f(x) - f(0)}{x - 0}$ ..........we are approaching from the left, here $\displaystyle f(x) = 2x + 2$

you will notice that these limits are not equal, thus the derivative $\displaystyle f'(0) = \lim_{x \to 0}\frac {f(x) - f(0)}{x - 0}$ does not exist
• Oct 11th 2007, 07:21 PM
Jhevon
Quote:

Originally Posted by Truthbetold
Determine whether he curve ha a tangent at the indicated point. If it does, give its slope. If not, explain why not?
f(x)=\left\{\begin{array}{rr}x&\ 2-2x-x^2 {if }x\ge0\\ -x&\2x-2{if } x > 0\end{array}\right. there is the messed up LaTex if anyone could/will/may :) etc, answer this. thank you!
The above is at x= 0. The second inequality is supposed to have an > or = sign.
Ah.

I couldn't get it. :( The LaTEx, I mean.
Here it is in math:
f(x)= 2-2x-x^2, x < 0
piecewise 2x+ 2, x > or = 0.

LOL, here is the LaTex code:

f(x) = \left \{ \begin{array}{lr} 2 - 2x - x^2 , & \mbox{ if } x \ge 0 \\ 2x + 2, & \mbox{ if } x < 0 \end{array}\right.

it yields:

$\displaystyle f(x) = \left \{ \begin{array}{lr} 2 - 2x - x^2 , & \mbox{ if } x \ge 0 \\ 2x + 2, & \mbox{ if } x < 0 \end{array}\right.$