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Math Help - Piecewise Differentiability

  1. #1
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    Piecewise Differentiability

    Let f be the piecewise defined function given by f(x)=ax+b x≤2. 2-x^2 2<x. Find a and b so that f is differentiable at x=2.
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    Quote Originally Posted by Velvet Love View Post
    Let f be the piecewise defined function given by f(x)=ax+b x≤2. 2-x^2 2<x. Find a and b so that f is differentiable at x=2.
    This will work if the two pieces are equal at x=2 (so that f is continuous) and if their derivatives are equal at x=2.

    Spoiler:
    We need to solve these two equations:

    1.) a\cdot 2+b=2-2^2 \implies 2a+b=-2
    2.) a=-2\cdot 2 \implies a=-4

    Plug a=-4 back into the first equation to find that b=6.

    So f(x)=\left\{\begin{array}{lr}-4x+6&:x\leq 2\\-x^2+2&:x>2\end{array}\right\}
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    I actually solved this problem a couple of minutes ago.
    Now i'm having trouble with this:
    Find the line which is perpendicular to the graph of f(x)=x^2+1 at x=4.



    You will need to use the definition to compute the derivative of f.
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    Quote Originally Posted by Velvet Love View Post
    Let f be the piecewise defined function given by f(x)=ax+b x≤2. 2-x^2 2<x. Find a and b so that f is differentiable at x=2.
    first, remember that if a function is differentiable at x = c , then it is continuous at x = c.

    secondly ...

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

    for your piecewise function ...

    f'(2)^- = f'(2)^+

    \lim_{x \to 2^-} \frac{(ax+b) - f(2)}{x-2} = \lim_{x \to 2^+} \frac{(2-x^2) - f(2)}{x-2}
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    Nevermind. I found that one out too.
    But this one is impossible :/

    Find the tangent line to the graph of f(x)=1/(x+4) at x=-3.



    You will need to use the definition of the derivative to find the slope of the tangent line.



    Put your answer in the form y=mx+b.
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  6. #6
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    Quote Originally Posted by Velvet Love View Post
    Nevermind. I found that one out too.
    But this one is impossible :/

    Find the tangent line to the graph of f(x)=1/(x+4) at x=-3.



    You will need to use the definition of the derivative to find the slope of the tangent line.



    Put your answer in the form y=mx+b.

    f'(-3) = \lim_{x \to -3} \frac{\frac{1}{x+4} - 1}{x - (-3)}

    start by combining \frac{1}{x+4} -1 into a single fraction.

    btw ... in future, start a new problem with a new thread.
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  7. #7
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Velvet Love View Post
    Nevermind. I found that one out too.
    But this one is impossible :/

    Find the tangent line to the graph of f(x)=1/(x+4) at x=-3.



    You will need to use the definition of the derivative to find the slope of the tangent line.



    Put your answer in the form y=mx+b.
    f'(x)=-\frac{1}{(x+4)^2}. Thus f'(-3)=-1. Therefore the slope of the perpendicular line will be 1, so you want the equation of the line with slope 1 going through the point (-3,1).

    y-1=1\cdot(x-(-3)) \implies \boxed{y=x+4}
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    Alright. I have ((1-x+4)/(x+4))/(x+3)
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  9. #9
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    Quote Originally Posted by Velvet Love View Post
    Alright. I have ((1 - x + 4)/(x+4))/(x+3)
    you have made a common algebra error ... try again.

    \frac{1}{x+4} - \frac{x+4}{x+4} = \frac{1-(x+4)}{x+4}<br />
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