Question

MAX AND MIN:


f(x) = x^3 - 3x + 1
i=[-3/2 , 3]

Answer 1

do u know calculus?

Answer 2

assuming u do

f'(x) =3 x^2 - 3 = 0 for critical points
x^2 = 1
x = 1 or -1
we now need to know which is max and which is a minm
f"(x) = 6x
when x = 1 f"(x) is +ve so this is minm
x = -1 is a maxm

f(x) maxm when f(x) = 3
minm at f(x) = -1

Answer 3

these are local maxm / minm values

Answer 4

that i represents an interval you're looking at for the function?

Answer 5

yes

Answer 6

i got max = 19

Answer 7

whenever you look for max/min on a closed interval you must evaluate at the enpoints in addition to the results posted by @cwrw238 .

did you get that 19 from the critical points?

Answer 8

test pts

Answer 9

how did you get the test points?

Answer 10

wait, yes i got 19 from my critical its

Answer 11

oh... ok... did you evaluate at x=-3/2 and x=3 also? because you need to compare that with the 19.

Answer 12

evaluate the function, not the derivative.

Answer 13

yes

Answer 14

because this is a closed interval, you should have a minimum also.

Answer 15

min = -1 at x=1

Answer 16

i guess you're done... good job. :)

Answer 17

another q- The sum of 2 nonnegative numbers is 36. Find these numbers if the 1st plus the square of the 2nd is a maximum

Answer 18

please help me.. my teacher is a awful explainer

Answer 19

HELP