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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 $\left(-\mathrm{\infty },1\right)$ and $\left(3,\mathrm{\infty }\right)$ and decreases in the interval (1,3). Given that ${P}^{\prime }\left(2\right)=0$ and $P\left(0\right)=4$, find P'(6).
In this problem, I have recognised that 2 is an inflection point and the derivative will be of the form: ${P}^{\prime }\left(x\right)=5\left(x-1\right)\left(x-3\right)\left(x-2\right)\left(x-\alpha \right)$.
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

Step 1
2 is an inflexion point so necessarily ${P}^{″}\left(2\right)=0$.
Step 2
Now ${P}^{\prime }\left(2\right)={P}^{″}\left(2\right)=0$ so 2 is a double root of P′.

Averi Fields

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

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