13 votes 13 votes Find the minimum value of $3-4x+2x^2$. Calculus gate1995 calculus maxima-minima easy descriptive + – Kathleen asked Oct 8, 2014 • recategorized Apr 25, 2021 by Lakshman Bhaiya Kathleen 2.6k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply skyby commented Dec 8, 2017 reply Follow Share (≤ 720) what is this? 0 votes 0 votes bhardwaj12345 commented Sep 21, 2020 reply Follow Share 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 votes 0 votes Please log in or register to add a comment.
Best answer 19 votes 19 votes $f(x) = 3-4x+2x^2$ $f '(x) = -4 + 4x = 0 \implies x=1$ $f ''(x) = 4$ $f ''(1) = 4>0$ Therefore at $x=1$ we will get minimum value, which is : $3 - 4(1) + 2(1)^2 = 1.$ jayendra answered Dec 27, 2014 • edited Apr 25, 2021 by Lakshman Bhaiya jayendra comment Share Follow See all 0 reply Please log in or register to add a comment.