A polynomial P(x) of degree 5 with lead coefficient one,increases in the ( − ∞ , 1 ) and ( 3 , ∞ ) and decreases in the interval (1,3)A polynomial function P(x) of degree 5 with leading coefficient one,increases in the interval ( − ∞ , 1 ) and ( 3 , ∞ ) and decreases in the interval (1,3). Given that P ( 0 ) = 4 and P ′ ( 2 ) = 0, find the value of P'(6).I noticed that 1,2,3 are the roots of the polynomial P′(x).P′(x) is a fourth degree polynomial. SoLet P ′ ( x ) = ( x − 1 ) ( x − 2 ) ( x − 3 ) ( a x + b ).Now I expanded P′(x) and integrated it to get P(x) and use the given condition P ( 0 ) = 4, but I am not able to calculate a, b.

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A polynomial P(x) of degree 5 with lead coefficient one,increases in the   (  −  ∞  ,  1  ) and   (  3  ,  ∞  ) and decreases in the interval (1,3)A polynomial function P(x) of degree 5 with leading coefficient one,increases in the interval   (  −  ∞  ,  1  ) and   (  3  ,  ∞  ) and decreases in the interval (1,3). Given that   P  (  0  )  =  4 and       P    ′    (  2  )  =  0, find the value of P&#039;(6).I noticed that 1,2,3 are the roots of the polynomial P′(x).P′(x) is a fourth degree polynomial. SoLet       P    ′    (  x  )  =  (  x  −  1  )  (  x  −  2  )  (  x  −  3  )  (  a  x  +  b  ).Now I expanded P′(x) and integrated it to get P(x) and use the given condition   P  (  0  )  =  4, but I am not able to calculate a, b.

Skylar Beard · Accepted Answer

Explanation:As   P  (  x  )  =      x    5    +  …, we have       P    ′    (  x  )  =  5      x    4    +  …. This gives us   a  =  5. Moreover, the facts that P is decreasing on (1,3) and       P    ′    (  2  )  =  0 imply that   x  =  2 must in fact be a multiple root of P′. We conclude that   2  a  +  b  =  0, so   b  =  −  10

A polynomial P(x) of degree 5 with lead coefficient one,increases in the (-infty, 1) and (3, infty) and decreases in the interval (1,3)

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