Question

Determine the extreme values of f(x)=1+(x+3)^2 on the the interval [-2,6]

Answer 1

check for stationary points on the interval

Answer 2

solve for f'(x)=0

Answer 3

solve f'(x) = 0 to find the stationary points,
check if it lies in the interval, if it lies in the interval, check for it's values,
and also check the values at the endpoints.

Answer 4

x=-3?

Answer 5

yeah ,,, but that's not in the interval ...so check the extreme values of the interval.

Answer 6

sorry for the wrong post
@experimentX is right

Answer 7

f'(-2)=-2
f'(6)= 18
?

Answer 8

no @siddharth18... that's the right procedure !!!

Answer 9

The answer is suppose to be
max: 82
min: 6

Answer 10

you maxima is 18 at x=6 and minima is -2 at x=-2

Answer 11

@experimentX i m talking about the one which i recently deleted

Answer 12

Oo .. looks like you put the value in f'(x)


don't put the value in f'(x) ... put it in f(x), then determinte the maxima and minima

Answer 13

i m feeling sleepy,so i am making silly mistakes..:(

Answer 14

yeah .. me too. it's bad to wake all night.
http://www.wolframalpha.com/input/?i=graph+f%28x%29%3D1%2B%28x%2B3%29^2+from+-2+to+6