Distance between local maximum and local minimum











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If $P (x)$ be a polynomial of degree $3$ satisfying $P (-1)=10,P (1)=-6$ and $P(x)$ has maximum at $x=-1$ and $P'(x)$ has minima at $x=1$.Find the distance between the local maximum and local minimum of the curve.










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    If $P (x)$ be a polynomial of degree $3$ satisfying $P (-1)=10,P (1)=-6$ and $P(x)$ has maximum at $x=-1$ and $P'(x)$ has minima at $x=1$.Find the distance between the local maximum and local minimum of the curve.










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      If $P (x)$ be a polynomial of degree $3$ satisfying $P (-1)=10,P (1)=-6$ and $P(x)$ has maximum at $x=-1$ and $P'(x)$ has minima at $x=1$.Find the distance between the local maximum and local minimum of the curve.










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      If $P (x)$ be a polynomial of degree $3$ satisfying $P (-1)=10,P (1)=-6$ and $P(x)$ has maximum at $x=-1$ and $P'(x)$ has minima at $x=1$.Find the distance between the local maximum and local minimum of the curve.







      euclidean-geometry maxima-minima






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      edited Nov 20 '16 at 12:33









      naveen dankal

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      asked Nov 20 '16 at 12:08









      user366398

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          Let the polynomial be $P(X)= ax^3+bx^2+cx+d$.



          As per the given conditions we have :



          $P(-1) = -a+b-c+d=10$



          $P(1) = a+b+c+d=-6$



          Also, $P'(-1)=3a-2b+c=0$



          And $P"(1) = 6a+2b=0$



          Solving for a,b,c,d gives



          $P(X)=x^3-3x^2-9x+5$



          => $P'(X)=3x^2-6x-9=3(x+1)(x-3)$



          This implies $x=-1$ is the point of maximum and $x=3$ is the point of minimum



          Hence, the distance between $(-1,10)$ and $(3,-22)$ is $4sqrt{65}$






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            1 Answer
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            1 Answer
            1






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            active

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            up vote
            2
            down vote



            accepted










            Let the polynomial be $P(X)= ax^3+bx^2+cx+d$.



            As per the given conditions we have :



            $P(-1) = -a+b-c+d=10$



            $P(1) = a+b+c+d=-6$



            Also, $P'(-1)=3a-2b+c=0$



            And $P"(1) = 6a+2b=0$



            Solving for a,b,c,d gives



            $P(X)=x^3-3x^2-9x+5$



            => $P'(X)=3x^2-6x-9=3(x+1)(x-3)$



            This implies $x=-1$ is the point of maximum and $x=3$ is the point of minimum



            Hence, the distance between $(-1,10)$ and $(3,-22)$ is $4sqrt{65}$






            share|cite|improve this answer



























              up vote
              2
              down vote



              accepted










              Let the polynomial be $P(X)= ax^3+bx^2+cx+d$.



              As per the given conditions we have :



              $P(-1) = -a+b-c+d=10$



              $P(1) = a+b+c+d=-6$



              Also, $P'(-1)=3a-2b+c=0$



              And $P"(1) = 6a+2b=0$



              Solving for a,b,c,d gives



              $P(X)=x^3-3x^2-9x+5$



              => $P'(X)=3x^2-6x-9=3(x+1)(x-3)$



              This implies $x=-1$ is the point of maximum and $x=3$ is the point of minimum



              Hence, the distance between $(-1,10)$ and $(3,-22)$ is $4sqrt{65}$






              share|cite|improve this answer

























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                Let the polynomial be $P(X)= ax^3+bx^2+cx+d$.



                As per the given conditions we have :



                $P(-1) = -a+b-c+d=10$



                $P(1) = a+b+c+d=-6$



                Also, $P'(-1)=3a-2b+c=0$



                And $P"(1) = 6a+2b=0$



                Solving for a,b,c,d gives



                $P(X)=x^3-3x^2-9x+5$



                => $P'(X)=3x^2-6x-9=3(x+1)(x-3)$



                This implies $x=-1$ is the point of maximum and $x=3$ is the point of minimum



                Hence, the distance between $(-1,10)$ and $(3,-22)$ is $4sqrt{65}$






                share|cite|improve this answer














                Let the polynomial be $P(X)= ax^3+bx^2+cx+d$.



                As per the given conditions we have :



                $P(-1) = -a+b-c+d=10$



                $P(1) = a+b+c+d=-6$



                Also, $P'(-1)=3a-2b+c=0$



                And $P"(1) = 6a+2b=0$



                Solving for a,b,c,d gives



                $P(X)=x^3-3x^2-9x+5$



                => $P'(X)=3x^2-6x-9=3(x+1)(x-3)$



                This implies $x=-1$ is the point of maximum and $x=3$ is the point of minimum



                Hence, the distance between $(-1,10)$ and $(3,-22)$ is $4sqrt{65}$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 20 '16 at 12:29

























                answered Nov 20 '16 at 12:17









                naveen dankal

                4,52321348




                4,52321348






























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