Point of negative inflection












1












$begingroup$



Find the range of values of a for which the function $f(x) = ax^3/3 +(a+2)x^2+(a-1)x + 2$ possesses a negative point of inflection.




My attempt



I differentiated it twice and equated it less than $0$ to get $x < frac{-(a+2)}{a}$. This isn't giving me any range of $a$. Any hint?






calculus







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  • $begingroup$
    Perhaps you should start by finding all the inflection points by finding the zeros of $f''(x)$
    $endgroup$
    – PossiblyDakota
    Sep 29 '18 at 5:23
















1












$begingroup$



Find the range of values of a for which the function $f(x) = ax^3/3 +(a+2)x^2+(a-1)x + 2$ possesses a negative point of inflection.




My attempt



I differentiated it twice and equated it less than $0$ to get $x < frac{-(a+2)}{a}$. This isn't giving me any range of $a$. Any hint?






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Perhaps you should start by finding all the inflection points by finding the zeros of $f''(x)$
    $endgroup$
    – PossiblyDakota
    Sep 29 '18 at 5:23














1












1








1





$begingroup$



Find the range of values of a for which the function $f(x) = ax^3/3 +(a+2)x^2+(a-1)x + 2$ possesses a negative point of inflection.




My attempt



I differentiated it twice and equated it less than $0$ to get $x < frac{-(a+2)}{a}$. This isn't giving me any range of $a$. Any hint?






calculus







share|cite|improve this question











$endgroup$





Find the range of values of a for which the function $f(x) = ax^3/3 +(a+2)x^2+(a-1)x + 2$ possesses a negative point of inflection.




My attempt



I differentiated it twice and equated it less than $0$ to get $x < frac{-(a+2)}{a}$. This isn't giving me any range of $a$. Any hint?






calculus




calculus






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share|cite|improve this question













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edited Feb 1 at 15:36







Zerix

















asked Sep 29 '18 at 5:16









ZerixZerix

1378




1378












  • $begingroup$
    Perhaps you should start by finding all the inflection points by finding the zeros of $f''(x)$
    $endgroup$
    – PossiblyDakota
    Sep 29 '18 at 5:23


















  • $begingroup$
    Perhaps you should start by finding all the inflection points by finding the zeros of $f''(x)$
    $endgroup$
    – PossiblyDakota
    Sep 29 '18 at 5:23
















$begingroup$
Perhaps you should start by finding all the inflection points by finding the zeros of $f''(x)$
$endgroup$
– PossiblyDakota
Sep 29 '18 at 5:23




$begingroup$
Perhaps you should start by finding all the inflection points by finding the zeros of $f''(x)$
$endgroup$
– PossiblyDakota
Sep 29 '18 at 5:23










1 Answer
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1












$begingroup$

$f(x) =frac{a}{3}x^3 +(a+2)x^2+(a-1) x+2$



I think you are supposed to find the intervals for which the function is concave-down? Note that the first derivative of the given function will allow you to find local mins and maxes:



$frac{df(x)}{dx} = ax^2 + 2(a + 2)x + (a-1) = 0$



$x = frac{-(a+2) +- sqrt{4a + 5}}{a}$



Check the concavity (sign of any point in $d^2 f/dx^2$) in the regions



$(-inf.,frac{-(a+2) - sqrt{4a + 5}}{a})$,



$(frac{-(a+2) - sqrt{4a + 5}}{a},frac{-(a+2) + sqrt{4a + 5}}{a})$,



$(frac{-(a+2) + sqrt{4a + 5}}{a}, inf)$



Second derivative is (as you found)



$frac{df^2(x)}{dx^2} = 2ax + 2(a + 2)$



Check the sign of that function when evaluated at any convenient point in those intervals. You want - .






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    $begingroup$

    $f(x) =frac{a}{3}x^3 +(a+2)x^2+(a-1) x+2$



    I think you are supposed to find the intervals for which the function is concave-down? Note that the first derivative of the given function will allow you to find local mins and maxes:



    $frac{df(x)}{dx} = ax^2 + 2(a + 2)x + (a-1) = 0$



    $x = frac{-(a+2) +- sqrt{4a + 5}}{a}$



    Check the concavity (sign of any point in $d^2 f/dx^2$) in the regions



    $(-inf.,frac{-(a+2) - sqrt{4a + 5}}{a})$,



    $(frac{-(a+2) - sqrt{4a + 5}}{a},frac{-(a+2) + sqrt{4a + 5}}{a})$,



    $(frac{-(a+2) + sqrt{4a + 5}}{a}, inf)$



    Second derivative is (as you found)



    $frac{df^2(x)}{dx^2} = 2ax + 2(a + 2)$



    Check the sign of that function when evaluated at any convenient point in those intervals. You want - .






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      $f(x) =frac{a}{3}x^3 +(a+2)x^2+(a-1) x+2$



      I think you are supposed to find the intervals for which the function is concave-down? Note that the first derivative of the given function will allow you to find local mins and maxes:



      $frac{df(x)}{dx} = ax^2 + 2(a + 2)x + (a-1) = 0$



      $x = frac{-(a+2) +- sqrt{4a + 5}}{a}$



      Check the concavity (sign of any point in $d^2 f/dx^2$) in the regions



      $(-inf.,frac{-(a+2) - sqrt{4a + 5}}{a})$,



      $(frac{-(a+2) - sqrt{4a + 5}}{a},frac{-(a+2) + sqrt{4a + 5}}{a})$,



      $(frac{-(a+2) + sqrt{4a + 5}}{a}, inf)$



      Second derivative is (as you found)



      $frac{df^2(x)}{dx^2} = 2ax + 2(a + 2)$



      Check the sign of that function when evaluated at any convenient point in those intervals. You want - .






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        $f(x) =frac{a}{3}x^3 +(a+2)x^2+(a-1) x+2$



        I think you are supposed to find the intervals for which the function is concave-down? Note that the first derivative of the given function will allow you to find local mins and maxes:



        $frac{df(x)}{dx} = ax^2 + 2(a + 2)x + (a-1) = 0$



        $x = frac{-(a+2) +- sqrt{4a + 5}}{a}$



        Check the concavity (sign of any point in $d^2 f/dx^2$) in the regions



        $(-inf.,frac{-(a+2) - sqrt{4a + 5}}{a})$,



        $(frac{-(a+2) - sqrt{4a + 5}}{a},frac{-(a+2) + sqrt{4a + 5}}{a})$,



        $(frac{-(a+2) + sqrt{4a + 5}}{a}, inf)$



        Second derivative is (as you found)



        $frac{df^2(x)}{dx^2} = 2ax + 2(a + 2)$



        Check the sign of that function when evaluated at any convenient point in those intervals. You want - .






        share|cite|improve this answer











        $endgroup$



        $f(x) =frac{a}{3}x^3 +(a+2)x^2+(a-1) x+2$



        I think you are supposed to find the intervals for which the function is concave-down? Note that the first derivative of the given function will allow you to find local mins and maxes:



        $frac{df(x)}{dx} = ax^2 + 2(a + 2)x + (a-1) = 0$



        $x = frac{-(a+2) +- sqrt{4a + 5}}{a}$



        Check the concavity (sign of any point in $d^2 f/dx^2$) in the regions



        $(-inf.,frac{-(a+2) - sqrt{4a + 5}}{a})$,



        $(frac{-(a+2) - sqrt{4a + 5}}{a},frac{-(a+2) + sqrt{4a + 5}}{a})$,



        $(frac{-(a+2) + sqrt{4a + 5}}{a}, inf)$



        Second derivative is (as you found)



        $frac{df^2(x)}{dx^2} = 2ax + 2(a + 2)$



        Check the sign of that function when evaluated at any convenient point in those intervals. You want - .







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Sep 29 '18 at 6:06

























        answered Sep 29 '18 at 5:36









        hatinacat2000hatinacat2000

        395




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