Minimum Value (1)

Calculus Level 3

What is the minimum value of the expression below?

f ( x ) = ( x − 2016 ) ( x − 2017 ) ( x − 2018 ) ( x − 2019 ) f(x)=(x-2016)(x-2017)(x-2018)(x-2019)

1 4 ( 2016 ) ( 2017 ( 2018 ) ( 2019 ) \frac{1}{4}(2016)(2017(2018)(2019) None of the others − 1 -1 ( 2016 ) ( 2017 ( 2018 ) ( 2019 ) (2016)(2017(2018)(2019) 1 1

Achmad Damanhuri by Achmad Damanhuri , 26, Indonesia

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1 solution

f ( x ) = ( x − 2016 ) ( x − 2017 ) ( x − 2018 ) ( x − 2019 ) Let x = u + 2017.5 g ( u ) = ( u + 3 2 ) ( u + 1 2 ) ( u − 1 2 ) ( u − 3 2 ) = ( u 2 − 9 4 ) ( u 2 − 1 4 ) = u 4 − 5 2 u 2 + 9 16 = ( u 2 − 5 4 ) 2 − 1 \begin{aligned} f(x) & = (x-2016)(x-2017)(x-2018)(x-2019) & \small \color{#3D99F6} \text{Let } x = u+2017.5 \\ g(u) & = \left(u+\frac 32\right) \left(u+\frac 12\right) \left(u-\frac 12\right) \left(u-\frac 32\right) \\ & = \left(u^2-\frac 94\right) \left(u^2-\frac 14\right) \\ & = u^4 - \frac 52 u^2 + \frac 9{16} \\ & = \left(u^2 - \frac 54\right)^2 - 1 \end{aligned}

Since ( u 2 − 5 4 ) 2 ≥ 0 \left(u^2 - \dfrac 54\right)^2 \ge 0 , min ⁡ ( f ( x ) ) = min ⁡ ( g ( u ) ) = − 1 \min(f(x)) = \min (g(u)) = \boxed{-1} .

0 Helpful 0 Interesting 6 Brilliant 0 Confused

This could be solved algebraically also. u^4-(5/2)u^2+(9/16)=(u^2-(5/4))^2-1. Therefore the minimum of the expression is -1

A Former Brilliant Member - 2 years, 2 months ago

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Thanks, I noticed that earlier but it is a Calculus problem. I will edit it..

Chew-Seong Cheong - 2 years, 2 months ago

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