# Math Help - Tangent of a Curve

1. ## 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!

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!

2. 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!

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

3. So no tangent at sharp turn, got it.
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.

4. 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.

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

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.

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

and

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

you will notice that these limits are not equal, thus the derivative $f'(0) = \lim_{x \to 0}\frac {f(x) - f(0)}{x - 0}$ does not exist

5. 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:

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