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4 votes
What is the least possible value of (x +1)(x+2)(x+3)(x +4)+2019 where x is a real

number?
MANY POINTS

2 Answers

3 votes

Answer:

f(x)=(x+1)(x+2)(x+3)(x+4)+2019

f(x)=(x2+5x+4)(x2+5x+6)+2019

Suppose that y=x2+5x

Hence we have f(y)f(y)=(y+4)(y+6)+2019=y2+10y+24+2019=y2+10y+25+2018=(y+5)2+2018≥2018[∵(y+5)2≥0,∀y∈R]

and therefore…. min (f(x))=2018

ANSWER = 2018

Explanation:

hope that helps >3

User Peter Hahndorf
by
7.5k points
4 votes

Answer:

2018

Explanation:

By grouping the first, last and two middle terms, we get (
x^(2)+5x+4)(
x^(2)+5x+6) + 2019. This can then be simplified to (
x^(2)+5x+2)^2 - 1 + 2019 Noting that squares are nonnegative, and verifying that
x^(2) + 5x + 5 = 0 for some real x, the answer is 2018.

User Cemron
by
7.4k points

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