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Thread: solving for a continuous function

  1. #1
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    solving for a continuous function

    let m and b be real numbers and let the function f be defined by
    $\displaystyle f(x)=\begin{cases}
    & \text{ if } x= 1+3bx+2x^2 (x\leq 1) \\
    & \text{ if } x= mx+b (x greater than 1)
    \end{cases}$
    If f is both continuous and differentiable at x=1, then
    m=?
    b=?

    my steps
    -substitute for x=1
    $\displaystyle 1+3b(1)+2(1)^2=m(1)+b$
    -combine like terms
    $\displaystyle 3+3b=m+b$
    -rearrange for m
    $\displaystyle m=3+2b$
    -substitute back in
    3+3b=(3+2b)+b
    3+3b=3+3b

    -this is where i am lost i dont even have a variable when I solve...
    what did I do wrong?

    All help is appreciated.
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  2. #2
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    Quote Originally Posted by dandaman View Post
    let m and b be real numbers and let the function f be defined by
    $\displaystyle f(x)=\begin{cases}
    & \text{ if } x= 1+3bx+2x^2 (x\leq 1) \\
    & \text{ if } x= mx+b (x greater than 1)
    \end{cases}$
    If f is both continuous and differentiable at x=1, then
    m=?
    b=?
    If $\displaystyle f$ is continous it means $\displaystyle \lim_{x\to 1^{-}} f(x) = \lim_{x\to 1^{+}} f(x)$.
    Thus, $\displaystyle \lim_{x\to 1^{-}} 1 + 3bx+2x^2 = \lim_{x\to 1^+} mx+b$. [1]
    If $\displaystyle f$ is differenciable it means $\displaystyle f'(1-) = f'(1+)$ (derivative from left equals derivative from right).
    Thus, $\displaystyle \frac{d}{dx}(1+3bx+2x^2)\bigg|_{x=1} = \frac{d}{dx}(mx+b)\bigg|_{x=1}$. [2]

    The equations [1] and [2] allow you to solve a system of equations.
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