Angelina508id
Answered

What is the least possible value of (x +1)(x+2)(x+3)(x +4)+2019 where x is a real
number?
MANY POINTS

Sagot :

Melodysmith14id

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

Step-by-step explanation:

hope that helps >3

Jreid123id

Answer:

2018

Step-by-step explanation:

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