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Math Help - Find the interval in which the function concave up??

  1. #1
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    Find the interval in which the function concave up??

    The graph of the function

    is concave up on the interval:

    (-infinity,infinity)

    None

    (1,infinity)


    (-infinity,0)U(1,infinity)

    (0,infinity)


    =======================

    I solve the question and my answer is :


    The graph is concave up on (-infinity,-0.5)U(0,infinity)

    and I'm sure from my answer but my answer is not included in the choices

    and When I use the graphics calculator the graph is concave up on all interval


    So what is the correct answer??
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  2. #2
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    You can use derivatives to answer this question

    First you must find the critical points, take your first equation and differentiate and solve for 0

    f(x)=x^4+x^3+x

    f\prime(x)=4x^3+3x^3+1

    Now that its differentiated solve for 0

    4x^3+3x^2+1=0

    x(4x^2+3x+1)=0

    you a critical point at x = -1.5 and x = 0

    now to find a graph is increasing you plug your critical points back in the derivative and if its <0 graph is decreasing >0 increasing

    to find concavity you take second derivative and plug in critical points if

    >0 then concave up and <0 concave down

    the second derivative is

    12x^2+6

    plus in 0 and -1.5

    plug 0 in and you see it is >0 so intervals from 0 to infinity is concave up

    plug -1.5 it is also concave up because > 0 so from -inf. to -1.5

    sorry for lack of work studying for finals
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  3. #3
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    Quote Originally Posted by sk8erboyla2004 View Post
    You can use derivatives to answer this question

    First you must find the critical points, take your first equation and differentiate and solve for 0

    f(x)=x^4+x^3+x

    f\prime(x)=4x^3+3x^3+1

    Now that its differentiated solve for 0

    4x^3+3x^2+1=0

    x(4x^2+3x+1)=0

    you a critical point at x = -1.5 and x = 0

    now to find a graph is increasing you plug your critical points back in the derivative and if its <0 graph is decreasing >0 increasing

    to find concavity you take second derivative and plug in critical points if

    >0 then concave up and <0 concave down

    the second derivative is

    12x^2+6

    plus in 0 and -1.5

    plug 0 in and you see it is >0 so intervals from 0 to infinity is concave up

    plug -1.5 it is also concave up because > 0 so from -inf. to -1.5

    sorry for lack of work studying for finals

    Thank you for replying..

    but I would to remind you that in concavity test we do the following :

    1)find first derivative

    2)find second derivative

    3)set first derivative=0 and solve for x

    4)usinig the zeros of second derivative we will set open interval and select test number within each interval to determine the type of concavity ..


    so your answer is false..
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  4. #4
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    Quote Originally Posted by change_for_better View Post
    The graph of the function

    is concave up on the interval:

    (-infinity,infinity)

    None

    (1,infinity)


    (-infinity,0)U(1,infinity)

    (0,infinity)


    =======================

    I solve the question and my answer is :


    The graph is concave up on (-infinity,-0.5)U(0,infinity)

    and I'm sure from my answer but my answer is not included in the choices

    and When I use the graphics calculator the graph is concave up on all interval


    So what is the correct answer??
    A function is concave upward when its second derivative is positive.
    If y= x^4+ x^3+ x, then y"= 12x^2+ 6x= 6x(2x+ 1). That is 0 at x= 0 and x= -1/2. If x< -1, x< 0 and 2x+1< 0 so y" is positive. If -1< x< 0, x< 0 and 2x+1> 0 so y" is negative. If x> 0, 2x+1> 0 so y" is positive. y is concave upward on exactly the interval you give, whether is one of the options or not!
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  5. #5
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    Quote Originally Posted by change_for_better View Post
    Thank you for replying..

    but I would to remind you that in concavity test we do the following :

    1)find first derivative

    2)find second derivative

    3)set first derivative=0 and solve for x

    4)usinig the zeros of second derivative we will set open interval and select test number within each interval to determine the type of concavity ..


    so your answer is false..
    If you know this why you couldnt have used this to answer your own question?
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    A function is concave upward when its second derivative is positive.
    If y= x^4+ x^3+ x, then y"= 12x^2+ 6x= 6x(2x+ 1). That is 0 at x= 0 and x= -1/2. If x< -1, x< 0 and 2x+1< 0 so y" is positive. If -1< x< 0, x< 0 and 2x+1> 0 so y" is negative. If x> 0, 2x+1> 0 so y" is positive. y is concave upward on exactly the interval you give, whether is one of the options or not!
    thank you for replying

    I would like to ask you ,when I use graphics calculator the shape of the graph of the function is concave up in all interval,butwhen we solve the question the interval (-0.5,0)is concave down ??
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  7. #7
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    Quote Originally Posted by sk8erboyla2004 View Post
    If you know this why you couldnt have used this to answer your own question?

    Because ,when I use graphics calculator the shape of the graph of the function is concave up in all interval,butwhen we solve the question the interval (-0.5,0)is concave down ??
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  8. #8
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    Hello, change_for_better!

    I agree with your intervals . . .


    The graph of the function: . f(x) \;=\;x^4+x^3+x is concave up on the interval:

    . . (a)\;(-\infty, \infty) \qquad(b)\text{ None} \qquad (c)\;(1,\infty) \qquad (d)\;(-\infty,0) \;\cup\; (1,\infty) \qquad(e)\;(0,\infty)

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    I solved and my answer is: . \left(-\infty,-\tfrac{1}{2}\right) \cup (0,\infty)

    and I'm sure from my answer but my answer is not included in the choices.


    and when I use the graphics calculator the graph is concave up on all interval
    . .
    Um ... not quite!



    So what is the correct answer?
    Your intervals are correct!


    \begin{array}{ccc}f(x) &=&x^4+x^3+x \\ f'(x) &=& 4x^3+3x^2+1 \\ f''(x) &=& 12x^2+6x \end{array}

    f''(x) > 0 \quad\Rightarrow\quad x < \text{-}\tfrac{1}{2}\:\text{ or }\:x > 0


    If you zoom in on your graph, there is a concave-down portion.
    Code:
                        |
                        |         * 
                        |        *
                        |      *
        *               |   *
      - - - - - - - - - o - - - - - - - - 
         *           o  |
          *        o    |
            *     *     |
               *        |
                        |

    And it's right where you predicted! . . . \left(\text{-}\tfrac{1}{2},\:0\right)

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  9. #9
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    Quote Originally Posted by Soroban View Post
    Hello, change_for_better!

    I agree with your intervals . . .

    Your intervals are correct!


    \begin{array}{ccc}f(x) &=&x^4+x^3+x \\ f'(x) &=& 4x^3+3x^2+1 \\ f''(x) &=& 12x^2+6x \end{array}

    f''(x) > 0 \quad\Rightarrow\quad x < \text{-}\tfrac{1}{2}\:\text{ or }\:x > 0


    If you zoom in on your graph, there is a concave-down portion.
    Code:
                        |
                        |         * 
                        |        *
                        |      *
        *               |   *
      - - - - - - - - - o - - - - - - - - 
         *           o  |
          *        o    |
            *     *     |
               *        |
                        |
    And it's right where you predicted! . . . \left(\text{-}\tfrac{1}{2},\:0\right)

    Thank you very very very very much Soroban for great explanation

    Now I understand my mistake ..

    So from choices I should select none because the correct answer is not included ..
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