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2022-09-15

Why is 2 double root of the derivative?

A polynomial function P(x) with degree 5 increases in the interval $(-\mathrm{\infty},1)$ and $(3,\mathrm{\infty})$ and decreases in the interval (1,3). Given that ${P}^{\prime}(2)=0$ 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: ${P}^{\prime}(x)=5(x-1)(x-3)(x-2)(x-\alpha )$.

But I am unable to understand why $x=2$ is double root of the derivative (i.e. why is $\alpha =2$?). It's not making sense to me. I need help with that part.

A polynomial function P(x) with degree 5 increases in the interval $(-\mathrm{\infty},1)$ and $(3,\mathrm{\infty})$ and decreases in the interval (1,3). Given that ${P}^{\prime}(2)=0$ 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: ${P}^{\prime}(x)=5(x-1)(x-3)(x-2)(x-\alpha )$.

But I am unable to understand why $x=2$ is double root of the derivative (i.e. why is $\alpha =2$?). It's not making sense to me. I need help with that part.

Harold Beltran

Beginner2022-09-16Added 3 answers

Step 1

2 is an inflexion point so necessarily ${P}^{\u2033}(2)=0$.

Step 2

Now ${P}^{\prime}(2)={P}^{\u2033}(2)=0$ so 2 is a double root of P′.

2 is an inflexion point so necessarily ${P}^{\u2033}(2)=0$.

Step 2

Now ${P}^{\prime}(2)={P}^{\u2033}(2)=0$ so 2 is a double root of P′.

Averi Fields

Beginner2022-09-17Added 3 answers

Explanation:

If 2 is an inflexion point then it is necessary that ${P}^{\u2033}(2)=0$, which means that you have a factor of $(x-2{)}^{2}$ in P′(x).

If 2 is an inflexion point then it is necessary that ${P}^{\u2033}(2)=0$, which means that you have a factor of $(x-2{)}^{2}$ in P′(x).

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