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Math Help - Differentiation and Need HELP with LATEX!

  1. #1
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    Unhappy Differentiation and Need HELP with LATEX!

    g(x) = {-x, x<=0
    3x^2,x>0

    (a) Evaluate the limit of {g(x+delta x) - g(x)}/{delta x) for x<=0 and x>0 as delta x tends to 0

    (b) evaluate the limit of {g(delta x) - g(0)}/{delta x) as delta x tends to 0 from the right and as delta x tends to 0 from the left

    (c) sketch the graph of g'(x)

    Is g(x) continuous at x=0?

    I need help with (b) and (c) only. Is there any user-friendly programmes to type the equations? I find the syntax of latex very difficult to learn. Btw, how do you type the above equations (using latex)?

    Thanks!
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  2. #2
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    B is simply asking you to compute the separate limits for the left and right side of the equation corresponding the respective piece of your piece-wise function. You are meant to see whether the limit from the right is the same as the limit from the left.
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  3. #3
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    Quote Originally Posted by ANDS! View Post
    B is simply asking you to compute the separate limits for the left and right side of the equation corresponding the respective piece of your piece-wise function. You are meant to see whether the limit from the right is the same as the limit from the left.
    Does this mean I must find the limit for -x,  x\leq 0 as
    delta x approaches 0 from the right and from the left as well as the limit for 3x^2 ,  x>0 as delta x approaches 0 from the right and from the left?

    Can anyone show me the working??? Thank you.
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  4. #4
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    Quote Originally Posted by cyt91 View Post
    Does this mean I must find the limit for -x,  x\leq 0 as
    delta x approaches 0 from the right and from the left as well as the limit for 3x^2 ,  x>0 as delta x approaches 0 from the right and from the left?

    Can anyone show me the working??? Thank you.
    For (a), you want to take x< 0 (not \le 0) and can assume that delta x is small enough that x+ delta x is also negative- then you can use "-x" as the formula for both g(x) and g(x+ delta x). If g(x)= -x, what is g(x+ delta x)?

    Then take x> 0 and assume that delta x is small enough that x+ delta x is also positive- then you can use " 3x^2 as the formula for both g(x) and g(x+ delta x). If g(x)= 3x^2, what is g(x+ deltax)?

    For (b), where x=0, taking the limit "from the left" means using g(0+ delta x)= -delta x and taking the limit "from the right" means using g(0+ delta x)= 3(delta x)^2.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    For (a), you want to take x< 0 (not \le 0) and can assume that delta x is small enough that x+ delta x is also negative- then you can use "-x" as the formula for both g(x) and g(x+ delta x). If g(x)= -x, what is g(x+ delta x)?

    Then take x> 0 and assume that delta x is small enough that x+ delta x is also positive- then you can use " 3x^2 as the formula for both g(x) and g(x+ delta x). If g(x)= 3x^2, what is g(x+ deltax)?

    For (b), where x=0, taking the limit "from the left" means using g(0+ delta x)= -delta x and taking the limit "from the right" means using g(0+ delta x)= 3(delta x)^2.
    For (a) :
    x<0,
    g(x+delta x)= -(x+delta x)
    Is this correct?
    Then, the limit for [g(x+delta x)-g(x)]/[delta x] as delta x tends to 0 is
    {-(x+delta x)-[-x]}/{delta x}=-1
    Correct?

    x>0,
    g(x+delta x)=3(x+delta x)^2
    Is this correct?
    Then the limit for [g(x+delta x)-g(x)]/[delta x] as delta x tends to 0 is
    {3(x+delta x)^2-3x^2}/[delta x]=6x
    Correct?

    For (b):
    when delta x approaches 0 from the left, the limit of
    {g(0+delta x)-g(0)}/{delta x} = {-(0+delta x)-[-0]}/[delta x]
    =-1

    and when delta x approaches 0 from the right, the limit of
    {g(0+delta x)-g(0)}/{delta x}
    = {3(0+delta x)^2-3(delta x)^2}/[delta x]
    =0

    Are my workings correct?

    How about the graph? Is it g'(x)=-1 for x<=0 and g'(x)=6x for x>0?

    Hence, is g(x) not continuous at x=0 since g'(x) for x<=0 and g'(x) for x>0 are not equal?

    Thank you!
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  6. #6
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    Quote Originally Posted by cyt91 View Post
    For (a) :
    x<0,
    g(x+delta x)= -(x+delta x)
    Is this correct?
    Then, the limit for [g(x+delta x)-g(x)]/[delta x] as delta x tends to 0 is
    {-(x+delta x)-[-x]}/{delta x}=-1
    Correct?
    Yes, if x< 0 and |\delta x|< |x|, which we can assume since we are taking the limit as \delta x goes to 0, x+ \delta x is also < 0 so f(x+ \delta x)- f(x)= -(x+\delta x)- x= -\delta x.

    x>0,
    g(x+delta x)=3(x+delta x)^2
    Is this correct?
    Yes, if x> 0 and \delta x> x , which we can assume since we are taking the limit as \delta x goes to 0, x+ \delta x is also > 0 so f(x+ \delta x)- f(x)= 3(x+ \delta x)^2- 3x^2.

    [quote]Then the limit for [g(x+delta x)-g(x)]/[delta x] as delta x tends to 0 is
    {3(x+delta x)^2-3x^2}/[delta x]=6x
    Correct?[/tex]
    Yes. 3(x+ \delta x)^2= 3(x^2+ 2x\delta x+ \delta x^2)= 3x^2- 6x\delta x+ \delta x^2[/tex]. 3(x+ \delta x)^2- 3x^2= 6x\delta+  \delta x^2 so (3(x+ \delta x)^2- 3x^2)/\delta x= 6x+ \delta x and the limit of that, as \delta x goes to 0 is 6x.

    Very good!

    For (b):
    when delta x approaches 0 from the left, the limit of
    {g(0+delta x)-g(0)}/{delta x} = {-(0+delta x)-[-0]}/[delta x]
    =-1

    and when delta x approaches 0 from the right, the limit of
    {g(0+delta x)-g(0)}/{delta x}
    = {3(0+delta x)^2-3(delta x)^2}/[delta x]
    =0
    Both correct.

    Are my workings correct?

    How about the graph? Is it g'(x)=-1 for x<=0 and g'(x)=6x for x>0?

    Hence, is g(x) not continuous at x=0 since g'(x) for x<=0 and g'(x) for x>0 are not equal?

    Thank you!
    No, "continuous at x= a" mean \lim_{x\to a} g(x)= g(a). It has nothing to do with g' because most continuous functions are not differentiable! And, in fact, to have a derivative, a function must be continuous.

    Here, the point is that \lim_{x\to 0^-} g(x)= \lim_{x\to 0} -x= 0 and \lim_{x\to 0^+} g(x)= \lim_{x\to 0} 3x^2= 0.

    So \lim_{x\to 0} g(x)= 0= g(0) and g is continuous at x= 0.
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