Why is 2 double root of the derivative?A polynomial function P(x) with degree 5 increases in the interval ( − ∞ , 1 ) and ( 3 , ∞ ) and decreases in the interval (1,3). Given that P ′ ( 2 ) = 0 and P ( 0 ) = 4, find P'(6).In this problem, I have recognised that 2 is an inflection point and the derivative will be of the form: P ′ ( x ) = 5 ( x − 1 ) ( x − 3 ) ( x − 2 ) ( x − α ).But I am unable to understand why x = 2 is double root of the derivative (i.e. why is α = 2?). It's not making sense to me. I need help with that part.

Question

Why is 2 double root of the derivative?A polynomial function P(x) with degree 5 increases in the interval   (  −  ∞  ,  1  ) and   (  3  ,  ∞  ) and decreases in the interval (1,3). Given that       P    ′    (  2  )  =  0 and   P  (  0  )  =  4, find P&#039;(6).In this problem, I have recognised that 2 is an inflection point and the derivative will be of the form:       P    ′    (  x  )  =  5  (  x  −  1  )  (  x  −  3  )  (  x  −  2  )  (  x  −  α  ).But I am unable to understand why   x  =  2 is double root of the derivative (i.e. why is   α  =  2?). It&#039;s not making sense to me. I need help with that part.

Harold Beltran · Accepted Answer

Step 12  is an inflexion point so necessarily       P    ″    (  2  )  =  0.Step 2Now       P    ′    (  2  )  =      P    ″    (  2  )  =  0 so 2 is a double root of P′.

Averi Fields · Accepted Answer

Explanation:If 2 is an inflexion point then it is necessary that       P    ″    (  2  )  =  0, which means that you have a factor of   (  x  −  2      )    2   in P′(x).

A polynomial function P(x) with degree 5 increases in the interval (-infty, 1) and (3, infty) and decreases in the interval (1,3). Given that P′(2)=0 and P(0)=4, find P′(6).

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