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Limits and continuity

  • 09-05-2006 2:31pm
    #1
    Registered Users, Registered Users 2 Posts: 9,579 ✭✭✭


    Hi There,

    Hope someone can help me out here.
    1st year College here and just one thing in maths I can't seem to figure out. Nothing in notes about it either.
    Its a question that comes up quite often.
    An example of the question is as follows:


    Show that the function f(x) =
    | x^3 ....for x <= 2 |
    | 3x-2 ....for -2 < x < 2 |
    | x^2 ....for x >= 2 |

    is continous but not differencitiable at x = +=2


    Right first part i find ok. I check for continuinity at x = 2^- and 2^+ to see if the limit of x->2 = f(2) which it does and same for the others when x is greater than 2. So basically proving this function is continous is not the problem.
    Problem is second part. How do you do it?

    What my guess is, to differenciate f(x) =

    f'(x) = 2x^2 ....for x < = 2
    3 ....for -2 < x < 2
    2x ....for x >=2

    so now I check for continunity at x = 2.

    Lefthand limit of 2, = f(1.999) = 3
    Righthand lmit = f(2.111) = 2*2.111 = 4.111
    LHL not_equal_to RHL so limit doesn't exisit and f(2) does not equal limit as limit doesn't exist so the function is continuous.

    For the derivative to be defined, the function must be continuous at x = 2 but the function isn't continous as proven above so thus its not differencitiable?

    Is this how you do it or am I totally wrong.

    Thanks.

    Donal/Webmonkey


Comments

  • Closed Accounts Posts: 1,475 ✭✭✭Son Goku


    I don't know the specifics of your course, so although this is right it mightn't be what they want.

    Anyway..........

    When you differentiated f (x) and got f ' (x), you obtained a function which gives you the derivative of f(x) at each point.

    However as you noticed, the left and right sides don't have the same limit, so there is not "well defined" value for f ' (x) at x = 2.

    Therefore the derivative f ' (x) is not defined at that point, i.e. f(x) is not differentiable at that point.

    You were basically correct.

    (Wait until you get to functions that are every where differentiable but continuous nowhere and vice versa.)

    Any questions just ask.


  • Registered Users, Registered Users 2 Posts: 9,579 ✭✭✭Webmonkey


    Deadly. Thank you.
    When i was checking net before i asked the question here for a solution i came across the non-continous differentiable functions. They sound strange..looking foward to seeing them over the years :D

    Thanks.


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