Question

How can you say that the function is maximum or minimum?

Answer 1

see the value of a.

Answer 2

-a = sad face.

+a = happy face.

Answer 3

@ saifoo sad face is maximum then happy face is minimum?did i get it right?

Answer 4

are you in calculus?

Answer 5

saifoo pls look at this and check if i got it right pls....
y=2+12x-x^3
=12-3x^2
=3(x^2-4)
= (x+2)(x-2)
x+2=0 x-2=0
x=-2 x=2
a=2
x,2;x=1
y=(x+2)(x-2)
=(3)(-1)
\[y \prime \]=-3

x.2;x=3
y=(3=2)(3-2)
\[y \prime \]=5

x=a;x=2
y=(2+2)(2-2)
\[y \prime \]=0
so x=2 at minimum?

Answer 6

@noxi yes

Answer 7

@ saifoo sad face is maximum then happy face is minimum?did i get it right?
Right.

Answer 8

saifoo will take it from here

Answer 9

ok, sheesha so where are you stuck now?

Answer 10

@noxi yes

Answer 11

pls. look atmy solution if its correct....

Answer 12

is that a calculus question? :O

Answer 13

@noxi yes

Answer 14

i'm confused bec. in our book (-2,-14) minimum,(2,18) maximum.then in my solution -2 is maximum then 2 is minimum.

Answer 15

@noxi yes

Answer 16

@ saifoo.yes.differential calculus

Answer 17

@noxi yes

Answer 18

2 saifoo and noxi.pls look at my solution pls...

Answer 19

Dang, sorry im learning that now a days! :/

Answer 20

@noxi yes

Answer 21

ok...tnx for trying.....

Answer 22

y=2+12x-x^3 is your question correCT/

Answer 23

then y' to me looks like 12-3x^2 and since you need the points of inflection which is your derivative =0

Answer 24

brother rolle said when you have 2 y values of different values there will a point between those two values where the derivative =0 . this means somewhere inbetween those two points theres a concave

Answer 25

ok , so 12-3x^2=0
i would get rid of a factor of 3
4-x^2=0
(2-x)(2+x)=0 looks good to me, yes?
x=2
x=-2
these are your critical points . now plug those into the original eq
f(2)=2+12(2)-(2)^3
f(2)=2+24-8=18
now try f(-2)
f(-2)=2-24-(-2)^3
f(-2)=-22+8=-14
at x=2 you have a max and at x=-2 you have a min
which is min (-2,-14) and max at (2,18)