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Math Help - first/second derivative test + inflection point

  1. #1
    Junior Member
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    first/second derivative test + inflection point

    Consider the function below.
    h(x) = (x + 1)^9 - 9x - 3

    (a)Find the intervals of increase.
    Find the interval of decrease.

    (b) Find the local minimum value.
    Find the local maximum value.

    (c) Find the inflection point.
    Find the interval the function is concave up.
    Find the interval the function is concave down.


    I know that
    (a) (-INFINITY, -2) U (0, INFINITY) for interval of increase
    (-2, 0) for interval of decrease

    (b) -2 for the local minimum value.
    14 for the local maximum value.

    (c) (-1, 6) for the inflection point.



    I got b(local minimum value) by finding the derivative which is
    h'(x) = 9(x + 1)^8 - 9
    then setting it equal to zero
    h'(x) = 9(x + 1)^8 - 9 = 0
    h'(x) = 9(x + 1)^8 = 9
    h'(x) = (x + 1)^8 = 1
    h'(x) = x^8 = 0
    x = 0
    then i plugged in 0 to the original equation
    h(x) = (0 + 1)^9 - 9(0) - 3
    h(x) = 1 - 3
    h(x) = -2 (local minimum value)

    I got b(local maximum value) by plugging in -2 to the original equation
    h(x) = (-2 + 1)^9 - 9(-2) - 3
    h(x) = (-1) - (-18) - 3
    h(x) = 17 - 3
    h(x) = 14 (local maximum value)

    I got c(inflection point) by getting the 2nd derivative of the original equation
    h''(x) = 72(x+1)^7
    then setting it equal to zero
    h''(x) = 72(x+1)^7 = 0
    h''(x) = (x+1)^7 = 0
    h''(x) = x = -1 (first point)
    then to get the second point, i plugged in -1(first point) to the original equation
    h(x) = (-1 + 1)^9 - 9(-1) - 3
    h(x) = 0 + 9 - 3
    h(x) = 6(second point)



    My question is, is my logic correct on the way i solved the part B and the first part of part C?

    also, how do i solve part A without graphing it on a calculator? is there a way to find the points without graphing it?

    and how do i find the rest of part C, which is
    "Find the interval the function is concave up.
    Find the interval the function is concave down."

    thanks!
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  2. #2
    Member
    Joined
    Sep 2009
    Posts
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    Well first of, if you know that the function is increasing all the way up to -2 and then decreasing, how do you conclude that -2 is a local minimum??

    h'(x) = (x + 1)^8 = 1
    h'(x) = x^8 = 0
    Well this is wrong. You can not subtract the 1 on the right side into the bracket.

    h'(x) = (x + 1)^8 = 1
    Since the power is even (x+1)=1 or (x+1)=-1

    Hope that helps with this part.

    Now for C.
    <br />
h''(x) = x = -1 (first point)
    then to get the second point, i plugged in -1(first point) to the original equation
    What do you mean the second point? as far as I can see there is only one inflection point.

    and how do i find the rest of part C, which is
    "Find the interval the function is concave up.
    Find the interval the function is concave down."
    Think a bit about this one, when the function is concave up it's derivative is increasing (where it is defined).
    What does the second derivative look like when the first one is increasing ?
    It is positive. Do you understand why?
    So on what intervals is the second derivative positive and where is it negative?
    how do i solve part A without graphing it on a calculator? is there a way to find the points without graphing it?
    You find out where the function is increasing or decreasing by taking the derivative,
    and find out where the derivative is positive or negative.

    Hope that helps.
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