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Math Help - Algorithm Sketching

  1. #1
    Senior Member
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    Algorithm Sketching

    I have questions in regards to:

    Use the algorithm for curve sketching to sketch:

    a) (2x+1)/(x-1)
    b) (2x)/(x^2-25)


    A) When I am figuring out the intervals of increase and decrease, I am supposed to use the first derivative test..
    f'(x)=-3/(x-1)^2
    and then find its critical numbers
    f'(x)=0 when /
    f'(x)=undefined when x=1
    and then use the critical numbers to find where it is negative or positive, decreasing or increasing.
    x<1 = decreasing
    x>1 = decreasing
    so f is always decreasing.

    However, when I find the concavities, I find that the CU and CD intervals are not related to the intervals of decreasing and increasing..
    f''(x)=6/(x-1)^3
    critical numbers f''(x)=0 when x = /
    f''(x)=undefined when x=1
    Then do the second derivative test to find the concavities
    x<0 = negative, concave down
    x>0= positive, concave up

    so when I say "f is always decreasing" above in my first derivative test, that doesn't match the second derivative results with CU and CD... I have a CU in there, but if concave up is positive, does it mean it has to increase? How can it increase if f is always decreasing.. OR are these completely different things? please clarify this for me.

    I am wondering this similar question (can a concave up can occur when f is ALWAYS decreasing? and can a concave down occur when f is ALWAYS increasing?) with regards to part B as well.

    Thanks for your help.
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  2. #2
    MHF Contributor

    Joined
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    Quote Originally Posted by skeske1234 View Post
    I have questions in regards to:

    Use the algorithm for curve sketching to sketch:

    a) (2x+1)/(x-1)
    b) (2x)/(x^2-25)


    A) When I am figuring out the intervals of increase and decrease, I am supposed to use the first derivative test..
    f'(x)=-3/(x-1)^2
    and then find its critical numbers
    f'(x)=0 when /
    f'(x)=undefined when x=1
    and then use the critical numbers to find where it is negative or positive, decreasing or increasing.
    x<1 = decreasing
    x>1 = decreasing
    so f is always decreasing.

    However, when I find the concavities, I find that the CU and CD intervals are not related to the intervals of decreasing and increasing..
    f''(x)=6/(x-1)^3
    critical numbers f''(x)=0 when x = /
    f''(x)=undefined when x=1
    Then do the second derivative test to find the concavities
    x<0 = negative, concave down
    x>0= positive, concave up

    so when I say "f is always decreasing" above in my first derivative test, that doesn't match the second derivative results with CU and CD... I have a CU in there, but if concave up is positive, does it mean it has to increase? How can it increase if f is always decreasing.. OR are these completely different things? please clarify this for me.

    I am wondering this similar question (can a concave up can occur when f is ALWAYS decreasing? and can a concave down occur when f is ALWAYS increasing?) with regards to part B as well.

    Thanks for your help.
    Yes, of course. For a simpler example, y= x^3 is always increasing but is concave down for x< 0 and concave up for x> 0. Conversely, y= -x^3 is always decreasing but is concave up for x< 0 and concave down for x> 0. "Concavity" does not have anything to do with "increasing" or "decreasing".
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