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    Differentiable

    Determine if each function is differentiable at x=1. If it is find the dirivative, if not explain y not.

    a) f(x) = { 3x-2 if x<1 and x^3 if x >=1

    b) f(x) = { 2x+1 if x<1 and x^2 if x>=1
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by learn18 View Post
    a) f(x) = { 3x-2 if x<1 and x^3 if x >=1
    We are differentiable as long as we are continuous. so we know f(x) is differentiable at x = 1 if it is continuous at x = 1. you can show this by finding the left and right hand limits. i will just show you the graph, so you can see it's continuos.

    as you see, f(x) is differentiable here.

    using f(x) = x^3
    => f ' (x) = 3x^2
    Attached Thumbnails Attached Thumbnails Differentiable-quest1.jpg  
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  3. #3
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    Quote Originally Posted by learn18 View Post
    Determine if each function is differentiable at x=1. If it is find the dirivative, if not explain y not.

    a) f(x) = { 3x-2 if x<1 and x^3 if x >=1

    b) f(x) = { 2x+1 if x<1 and x^2 if x>=1
    Because these functions are both piecewise differentiable all we need to
    show is that:

    lim_{x->1-} f(x) = lim_{x->1+} f(x),

    and that:

    lim_{x->1-} f'(x) = lim_{x->1+} f'(x),

    If both of these hold then the function is differentiable at x=1, otherwise they
    are not.

    For a) lim_{x->1-} f(x) = 1, lim_{x->1+} f(x) = 1, and lim_{x->1-} f'(x) = 3,
    lim_{x->1+} f'(x) = 3, so f is differentiable at x=1.

    For b) lim_{x->1-} f(x) = 3, lim_{x->1+} f(x) = 1, so f is not differentiable
    at x=1.

    RonL
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by learn18 View Post
    Determine if each function is differentiable at x=1. If it is find the dirivative, if not explain y not.
    b) f(x) = { 2x+1 if x<1 and x^2 if x>=1
    See the graph below. Obviously f(x) is not continuous at the desired point, so we cannot find the derivative. You can show this by showing that the left and right hand limits are different (do you know how to do that?)

    EDIT: Nevermind, CaptainBlack did it for you
    Attached Thumbnails Attached Thumbnails Differentiable-quest2.jpg  
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Because these functions are both piecewise differentiable all we need to
    show is that:

    lim_{x->1-} f(x) = lim_{x->1+} f(x),

    and that:

    lim_{x->1-} f'(x) = lim_{x->1+} f'(x),

    If both of these hold then the function is differentiable at x=1, otherwise they
    are not.

    For a) lim_{x->1-} f(x) = 1, lim_{x->1+} f(x) = 1, and lim_{x->1-} f'(x) = 3,
    lim_{x->1+} f'(x) = 3, so f is differentiable at x=1.

    For b) lim_{x->1-} f(x) = 3, lim_{x->1+} f(x) = 1, so f is not differentiable
    at x=1.

    RonL
    I was not aware that we have to check the left and right hand limits for f', i thought once it checked out for f we can assume it would check out for f'. is this an incorrect assumption? can you give a counter example, where f is continous at the point but f' isn't

    EDIT: nevermind, learn18 provided us with a counter example in a subsequent problem, see http://www.mathhelpforum.com/math-he...erntiable.html
    Last edited by Jhevon; April 22nd 2007 at 01:13 PM.
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