Question

Extremal points, local/global maxima/minima
f(x) = x^2 + px + q, x E (-infinity, infinity)

Answer 1

I first made f'(x) = 0 so.. 2x+ p = 0, so x = -1/2 p but I have no idea what to do next :S help me please

Answer 2

f''(x) = 2 so its a minimum, but how do I tell if its global or local?????

Answer 3

minima which is smallest among all minimas then its global

Answer 4

but how did you know it was global?

Answer 5

have u find all the minimas of the equation

Answer 6

All I have really done is do the 2nd Derivation to find if it was maxima or minima but I don't know how to work out if it is local or global

Answer 7

if there is only one minima then its global ,

Answer 8

local minima is a relative term

Answer 9

I don't know if there is only one minima, thats what I really need help with :S

Answer 10

:)its tricky

Answer 11

so for f(x) = x^2 + px + q, I make f'(x) = 0 so x = -1/2p, what do I do after that??

Answer 12

dont know for diffrent values of x we will have diffrent minima ,as x depends on p ,

Answer 13

f(x) = x^2 +px +q is an upward parabola ---> it has minimum point only, right?

Answer 14

yes u r right bt the parabola cn be anywhere depends on the value of discremenent

Answer 15

and its vertex (-p/2, q) ok?

Answer 16

if p, q are arbitrary, so?? just argue on them. We don't need a special points to argue on, right?

Answer 17

The question is about local/ global max, min
Our parabola is upward, then we have min only,

Answer 18

the form of the parabola shows that it doesn't have local, then that min is global

Answer 19

we have min bt there cn be a no of minimas

Answer 20

and the parabola gets global min at -p/2 and its value is q
And we are done.
That's all I know. :)

Answer 21

sorry for being slow, but can you explain why its value is q? :S

Answer 22

the form \(y = ax^2+bx+c\) has vertex at \(x =\dfrac{-b}{2a}\)
to this problem \(x = \dfrac{-p}{2}\) , then plug it back to find y of the vertex
\((\dfrac{-p}{2})^2 +p*(\dfrac{-p}{2})+q\)
\(\dfrac{p^2}{4}-\dfrac{p^2}{4}+q=q\)
so that vertex is (p/2,q)

Answer 23

oh, I am sorry, mistake :)

Answer 24

isnt it p^2/4 - p^2/2 ?

Answer 25

yes, I am sooooo sorry for my carelessness. :)

Answer 26

no problem I appreciate you helping :)

Answer 27

But ganeshie8 is here, he can handle it. right? @ganeshie8 rescue me, pleeeeeease

Answer 28

would it be, global minimum at point (-1/2p , q - 3/4p^2)?

Answer 29

try this instead
http://www.wolframalpha.com/input/?i=min+and+max+x%5E2+%2B+c*x+%2B+d

Answer 30

could you please explain how to tell if its global or local please?

Answer 31

it is a quadratic, so you can have atmost one extrema

Answer 32

since the leading coefficient is positive its a upward facing parabola

Answer 33

so thats the only extrema and it is global too

Answer 34

makes sense, thank you :) , another example is min max, local or global for xlnx where x E [1,10] is that the sample idea as the example above?

Answer 35

yes, but u need to be careful when you're given an interval

Answer 36

the extrema can occur at boundary also

Answer 37

you need to evaluate the function at x = 1 and x = 10 also and compare them against other values at critical points

Answer 38

thats for global min/max ^^

Answer 39

so xlnx will have a min and max? because its not quadratic? will there be local and global ones also?

Answer 40

we don't know that yet, lets find the critical points by solving f' = 0

Answer 41

find the derivative, set it equal to 0 and solve x

Answer 42

f'(x) = 1 + ln(x) so 1+ln(x) = 0, so x = 1/e ?

Answer 43

yup! so ur critical x values are :

x = 1/e
x = 1
x = 10

Answer 44

so f(1/e) = 1/e * ln(1/e) = -1/e?

Answer 45

yes! evaluate the function at boundary values too

Answer 46

x = 1/e, y = -1/e
x = 1, y = ?
x = 10, y = ?

Answer 47

oh I see
one sec