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Math Help - How do I make this continuous and differentiable?

  1. #1
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    How do I make this continuous and differentiable?



    I'm lost since "a" and "b" are in the same formula

    But I think if I take the derivative of the first one, I might be able to figure out "a" in the second.

    Then I would just need a y value to find b.
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  2. #2
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    Quote Originally Posted by Zanderist View Post


    I'm lost since "a" and "b" are in the same formula

    But I think if I take the derivative of the first one, I might be able to figure out "a" in the second.

    Then I would just need a y value to find b.
    you're on the right track ...

    note that the function has to be continuous, and the derivative also has to be continuous.

    \lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x)  = -2a+b

    \lim_{x \to -2^-} f'(x) = \lim_{x \to -2^+} f'(x)  = a
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  3. #3
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    Hello, Zanderist!

    Find a and b so that the function: . f(x) \;=\;\bigg\{\begin{array}{ccc}4x^2-6x^2+8 && x < -2 \\ ax + b && x \geq -2 \end{array}
    . . is both continuous and differentiable.

    To be continuous, f(\text{-}2) be equal on "both branches".

    . . f(\text{-}2) \:=\:4(\text{-}2)^3 - 6(\text{-}|2)^2 + 8 \:=\:-44

    . . f(\text{-}2) \:=\:a(\text{-}2) + b \;=\;-2a + b

    . . Hence: . -2a + b \:=\:-44 .[1]


    To be differentiable, f'(\text{-}2) must be equal on "both branches."

    . . f'(x) \:=\:12x^2-12x \quad\Rightarrow\quad f'(\text{-}2) \:=\:12(\text{-}2)^2 - 12(\text{-}2) \:=\:72

    . . f'(x) \:=\:a

    . . Hence: . \boxed{a \:=\:72}


    Substitute into [1]: . -2(72) + b \:=\:-44 \quad\Rightarrow\quad\boxed{b \:=\:100}

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, Zanderist!


    To be continuous, f(\text{-}2) be equal on "both branches".

    . . f(\text{-}2) \:=\:4(\text{-}2)^3 - 6(\text{-}|2)^2 + 8 \:=\:-44

    . . f(\text{-}2) \:=\:a(\text{-}2) + b \;=\;-2a + b

    . . Hence: . -2a + b \:=\:-44 .[1]


    To be differentiable, f'(\text{-}2) must be equal on "both branches."

    . . f'(x) \:=\:12x^2-12x \quad\Rightarrow\quad f'(\text{-}2) \:=\:12(\text{-}2)^2 - 12(\text{-}2) \:=\:72

    . . f'(x) \:=\:a

    . . Hence: . \boxed{a \:=\:72}


    Substitute into [1]: . -2(72) + b \:=\:-44 \quad\Rightarrow\quad\boxed{b \:=\:100}


    Your right about "a" but it tells me that "b" (100) is the wrong answer.

    I've figured it out, that 44 is supposed to be -48.

    Thus meaning b = 96
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    By the way, if f is differentiable at, say, x= a, the derivative is not necessarily continuous there, but it does satisfy the "intermediate value property". That's why we can say that we must have \lim_{x\to a^+}f'(x)= \lim_{x\to a^-}f'(x).
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