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Math Help - Calculus (Limits and Derivatives)

  1. #1
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    Calculus (Limits and Derivatives)

    1. lim_{x\rightarrow-5}\frac{x^3+125}{(-5-x)^3}

    I reduced it to:

    lim_{x\rightarrow-5}\frac{x^{2}+5x+25}{x^{2}+10x+25}

    But I don't know what to do after that.





    2. Find equations of all tangent lines to the graph of y=4x^{3}+5x-8

    I took the derivative of the equation, which was:

    \acute{y}=12x^{2}+5

    I then thought I might need to make the derivative equal to the original equation, so:

    12x^{2}+5=4x^{3}+5x-8

    But I don't really know...




    3. Determine values of constant real numbers a, b so that the function:

    f(x)= ax^{2}-8x+6 if x\leq-1
    f(x)= bx+2 if x is greater than -1

    is differentiable at x=-1.

    No idea how to do this one.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by Thomas View Post
    1. lim_{x\rightarrow-5}\frac{x^3+125}{(-5-x)^3}

    I reduced it to:

    lim_{x\rightarrow-5}\frac{x^{2}+5x+25}{x^{2}+10x+25}

    But I don't know what to do after that.
    \displaystyle\lim_{x\to-5}\frac{x^3+125}{(-5-x)^3}=-\lim_{x\to-5}\frac{(x+5)(x^2-10x+25)}{(5+x)^3}=
    \displaystyle =-\lim_{x\to-5}\frac{x^2-10x+25}{(x+5)^2}=\frac{-100}{+0}=-\infty
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  3. #3
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    Ahh, I didn't think about how the bottom is squared.

    Thanks!
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  4. #4
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by Thomas View Post
    3. Determine values of constant real numbers a, b so that the function:

    f(x)= ax^{2}-8x+6 if x\leq-1
    f(x)= bx+2 if x is greater than -1

    is differentiable at x=-1.

    No idea how to do this one.
    First of all, f must be continuous at x=-1.
    \displaystyle\lim_{x\nearrow-1}f(x)=\lim_{x\nearrow-1}(ax^2-8x+6)=a+14=f(-1)
    \displaystyle\lim_{x\searrow-1}f(x)=\lim_{x\searrow-1}(bx+2)=-b+2.
    Then a+14=-b+2\Rightarrow a+b=-12 (1)
    We have \displaystyle f'(x)=\left\{\begin{array}{cc}2ax-8, & x\leq-1\\b, & x>-1\end{array}\right.

    f_l'(-1)=\lim_{x\nearrow-1}f'(x)=-2a-8, \ f_r'(-1)=\lim_{x\searrow-1}f'(x)=b

    f_l'(-1)=f_r'(-1)\Rightarrow-2a-8=b\Rightarrow 2a+b=-8 (2)

    Now, solve the system formed by equations (1) and (2).
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