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    Can you please check my answers? Thanks

    6. Compute f'(a) in two ways, using Eq. 1 and 2.

    Eq. 1:f'(a)=\lim_{h\to 0} \dfrac{f(a+h) - f(a)}{h}

    Eq. 2:f'(a)=\lim_{x\to a} \dfrac{f(x) - f(a)}{x-a}

    Where, x=a+h

    f(x)= 9 - 3x^2 , a = 0

    f'(0) = \lim_{h\to 0} \dfrac{f(0+h) - f(0)}{h}

    f'(0) = \lim_{h\to 0} \dfrac{f(h) - 9}{h}

    f'(0) = \lim_{h\to 0} \dfrac{9 - 3h^2 - 9}{h}

    f'(0) = \lim_{h\to 0} \dfrac{-3h^2}{h} = \lim_{h\to 0} -3h = 0


    f'(0) = \lim_{x\to a} \dfrac{f(x) - f(0)}{x-0}

    f'(0) = \lim_{x\to a} \dfrac{9 - 3x^2 - 9}{x} = \lim_{x\to a} \dfrac{-3x^2}{x} = \lim_{x\to a} -3x = 0.



    8. Refer to f(x) in figure 11 when answering this question.



    Compute \dfrac{f(2.5) - f(1)}{2.5 - 1}. What does this quantity represent?

    Answer:
    \dfrac{2-1}{1.5} = 2/3

    This quantity represents the average rate of change of y with respect to x over the interval [1, 2.5] (or the average rate of change of function f over the interval [1, 2.5]).



    22. First find the slope and then an equation of the tangent line to the graph of f(x) = \sqrt{x} at x=4 using the definition of a derivative.

    f'(4) = \lim_{h\to 0} \dfrac{f(4+h) - f(4)}{h}

    f'(4) = \lim_{h\to 0} \dfrac{\sqrt{4+h} - 2}{h}

    f'(4) = \lim_{h\to 0} \dfrac{\sqrt{4+h} - 2}{h} * \dfrac{\sqrt{4+h} + 2}{\sqrt{4+h} + 2}

    f'(4) = \lim_{h\to 0} \dfrac{(4+h) - 4}{h(\sqrt{4+h} + 2)} = \lim_{h\to 0} \dfrac{1}{\sqrt{4+h} + 2} = \dfrac{1}{4}

    Slope of tangent = 1/4.

    Point on tangent: (4, 2)

    Would I have to consider the tangent to the curve at (4, -2) as well?

    Equation of tangent:  y - 2 = 1/4 * (x - 4)
     y= x/4 + 1

    God, limits latex is a pain in the butt. :mad:

    Thanks again!
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    They all seem right to me.

    For Q22, you don't need to consider (4,-2) since \sqrt{x} refers only to the positive square root (otherwise it wouldn't be a function).
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    Using the limit definition, compute f'(x)

    f(x) = x^{\dfrac{-1}{2}}

    I'm having problems with the algebra.
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    (Original post by Doughboy)
    Using the limit definition, compute f'(x)

    f(x) = x^{\frac{-1}{2}}

    I'm having problems with the algebra.
    Using method 1 from your first post, once you've sorted the fractions in the numerator so everything is over a common denominator, what can you multiply top and bottom by to get rid of the square roots on top?

    Spoiler:
    Show


    You want to get to the form a^2-b^2 in the numerator.



    If you're still stuck, post where you've got to.
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    Hey, sorry but I figured out how to solve the problem.

    I think I ended up with a term including 2x\sqrt{x}

    I was having a hard time getting it to 2x^{3/2} because I was overlooking the simple laws of indices.

    Thanks
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    (Original post by Doughboy)
    Can you please check my answers? Thanks

    6. Compute f'(a) in two ways, using Eq. 1 and 2.

    Eq. 1:f'(a)=\lim_{h\to 0} \dfrac{f(a+h) - f(a)}{h}

    Eq. 2:f'(a)=\lim_{x\to a} \dfrac{f(x) - f(a)}{x-a}

    Where, x=a+h

    f(x)= 9 - 3x^2 , a = 0

    f'(0) = \lim_{h\to 0} \dfrac{f(0+h) - f(0)}{h}

    f'(0) = \lim_{h\to 0} \dfrac{f(h) - 9}{h}

    f'(0) = \lim_{h\to 0} \dfrac{9 - 3h^2 - 9}{h}

    f'(0) = \lim_{h\to 0} \dfrac{-3h^2}{h} = \lim_{h\to 0} -3h = 0


    f'(0) = \lim_{x\to a} \dfrac{f(x) - f(0)}{x-0}

    f'(0) = \lim_{x\to a} \dfrac{9 - 3x^2 - 9}{x} = \lim_{x\to a} \dfrac{-3x^2}{x} = \lim_{x\to a} -3x = 0.
    I'd possibly add another step in the last line for clarity: \lim_{x\to a} -3x =-3a=0 (as you have been very detailed so far in your working) But it shouldn't cost you any marks as written.
 
 
 
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