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Math Help - Various Calculus Problems (rate of change, piecewise)

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    Various Calculus Problems (rate of change, piecewise)

    Having some difficulty with these problems. I already have the answer to the second in the back of the book, but just can't figure out how to get there.

    1) Sketch the graph of a function f such that f' > 0 for all x and the rate of change of the function is decreasing.
    2) f(x) = ax^3 when x <= 2
    f(x) = ax^2 when x > 2
    Find a and b such that f is differentiable everywhere.
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    Quote Originally Posted by seuzy13 View Post
    Having some difficulty with these problems. I already have the answer to the second in the back of the book, but just can't figure out how to get there.

    1) Sketch the graph of a function f such that f' > 0 for all x and the rate of change of the function is decreasing.
    The fact that f'> 0 tells you the graph is increasing- draw a graph that is going up as you go to the right. "The rate of change of the function is decrasing" tells you that how fast it goes up is slowing. Draw a graph that is going up but less and less steep.

    2) f(x) = ax^3 when x <= 2
    f(x) = ax^2 when x > 2
    Find a and b such that f is differentiable everywhere.
    Obviously (I hope it is obvious!) both ax^3 and ax^2 are differentiable so the only problem is at the point x= 2.

    What is the derivative of ax^3 at x= 2? What is the derivative of ax^2 at x= 2? What value of x makes those the same?
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    Quote Originally Posted by seuzy13 View Post
    Having some difficulty with these problems. I already have the answer to the second in the back of the book, but just can't figure out how to get there.

    1) Sketch the graph of a function f such that f' > 0 for all x and the rate of change of the function is decreasing.

    sketch a graph that is increasing at a decreasing rate

    2) f(x) = ax^3 when x <= 2
    f(x) = ax^2 when x > 2
    Find a and b such that f is differentiable everywhere.

    sure you've posted f(x) correctly? where is "b" ?
    ...
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    So sorry, I goofed on the second part of 2) you're right.
    f(x) = x^2 + b when x > 2
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    Quote Originally Posted by seuzy13 View Post
    So sorry, I goofed on the second part of 2) you're right.
    f(x) = x^2 + b when x > 2
    note that for a function has to be continuous to be differentiable ...

    if f(x) is continuous at x = 2, then ...

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


    in addition, if f(x) is differentiable, then ...

    \lim_{x\to 2^-} f'(x) = \lim_{x\to 2^+} f'(x) = f'(2)
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    Could you give me a little more help please? I'm still stumped.
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    Quote Originally Posted by seuzy13 View Post
    Could you give me a little more help please? I'm still stumped.
    \lim_{x \to 2^-} ax^3 = ?

    \lim_{x \to 2^+} x^2+b = ?

    set the left limit = right limit ... that will be equation #1 with two unknowns, a and b.


    now do the same for f'(x) to get a second equation.
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    It's just not falling into place for some reason.
    First equation:
    8a = 4 + b
    Second equation:
    12a = 4 + b
    Which gives a = 0. Clearly incorrect.

    I think my problem is the second equation.
    <br />
\lim_{x\to 2^-} f'(x) = \lim_{x\to 2^+} f'(x) = f'(2)<br />
    3ax^2 = 2x + b = 12a
    Correct? I thought the derivative of b would equal zero there, but that would not help any.
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    Quote Originally Posted by seuzy13 View Post
    It's just not falling into place for some reason.
    First equation:
    8a = 4 + b
    Second equation:
    12a = 4 + b
    Which gives a = 0. Clearly incorrect.

    I think my problem is the second equation.
    <br />
\lim_{x\to 2^-} f'(x) = \lim_{x\to 2^+} f'(x) = f'(2)<br />
    3ax^2 = 2x + b = 12a
    Correct? I thought the derivative of b would equal zero there, but that would not help any.
    what is the derivative of x^2 + b ?
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  10. #10
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    2x...OHHHHhhh. Haha. I got it now. Thanks so much for your help.
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