Question

Please Help! ,Locate the extrema of the closed interval:
f(x)=-x^2+3x [0,3]

Answer 1

have you tried taking the first derivative yet?

Answer 2

yeah

Answer 3

I found the critical point which is 3/2

Answer 4

... I think

Answer 5

ok, so the extrema is when x=3/2

substitute that back into the original function to determine f(3/2)

Answer 6

so would it be 6.75?

Answer 7

9/2-9/4
18/4 - 9/4
9/4
2 1/4

Answer 8

how did you get 6.75?

Answer 9

isn't a negative times a negative a positive?

Answer 10

\[f(x)= -(x^2)+3x\]

Answer 11

always assume its like this, unless its written like this \[f(X) = (-x)^2+3x\]

Answer 12

oh yeah because its like -1(x^2) right?

Answer 13

yup

Answer 14

oh ok, that was messing me up, Thanks!

Answer 15

anyways, once you figure out f(3/2)

then the location of your extrema should be

(3/2,9/4)

Answer 16

yeah I understand now..