Results 1 to 5 of 5

Math Help - Tangent of a Curve

  1. #1
    Member
    Joined
    Oct 2007
    Posts
    178

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Truthbetold View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2007
    Posts
    178
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Truthbetold View Post
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Truthbetold View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Tangent on a curve
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 30th 2010, 05:32 AM
  2. Replies: 6
    Last Post: April 7th 2010, 02:34 PM
  3. tangent to curve
    Posted in the Calculus Forum
    Replies: 7
    Last Post: August 6th 2009, 05:19 PM
  4. Tangent Curve to a Curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 25th 2009, 12:03 AM
  5. Tangent to the curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 20th 2008, 04:25 AM

Search Tags


/mathhelpforum @mathhelpforum