in Calculus recategorized by
2,565 views
13 votes
13 votes
Find the minimum value of $3-4x+2x^2$.
in Calculus recategorized by
2.6k views

2 Comments

(≤ 720)

what is this?

0
0

2nd part of this question ….

Determine the number of positive integer(<=720)

      which are not divisible by any of 2,3, or 5.

solution:

total number of positive integers (<=720)=720 

no. of integers divisible by 2=floor(720/2)=360

no. of integers divisible by 3=floor(720/3)=240

no. of integers divisible by 5=floor(720/5)=144

Here we are going to see the formula for (A U B U C).

n(AUBUC)=n(A)+n(B)+n(C)-n(AnB)-n(BnC)-n(CnA)+n(AnBnC)

so number of positive integers(<720) which are  divisible by any of 2,3,or 5:

     =floor(720/2)+floor(720/3)+floor(720/5) – floor(720/6) – floor(720/15) –         floor(720/10).+floor(720/30)

      = 360+240+144 -120 -48-72+24

      = 528  positive integers numbers are there which are divisible by 2 ,3 or 5. between 1 to 720.

SO answer will be 720-528=192 numbers which are not divisible by  2 ,3 or 5.

 

 

 

0
0

1 Answer

19 votes
19 votes
Best answer
$f(x) = 3-4x+2x^2$

$f '(x) = -4 + 4x = 0  \implies  x=1$

$f ''(x) = 4$

$f ''(1) = 4>0$

 Therefore at $x=1$ we will get minimum value, which is : $3 - 4(1) + 2(1)^2 = 1.$
edited by
Answer:
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true