Question

the minimum value of f(x)=x^3/3-x^2-3x on [-2,2]?

Answer 1

Do i use the Extreme Value Theorem?

Answer 2

if i do the derivative of f(x) ; its x^2 - x -3 . Then what?

Answer 3

equate it to 0, and evaluate with respect to interval [-2,2]

Answer 4

the resulting value of x I mean as you equate f'(x)=0

Answer 5

so derive again right? so it becomes 2x-1. then?

Answer 6

no we solve for possible minimum points on interval [-2,2]...
f'(x)=0\[x^2-x-3=0\]\[(x+1)(x-2)=0\]therefore our minimum points are -1 and 2... are they in interval [-2,2]?

Answer 7

[-1,2] is inside [-2,2] so yes? so what next?

Answer 8

if they are... they are the minimum value of f(x) you'r looking for...

Answer 9

to find the y-ordinate, plug-in x at f(x)...

Answer 10

the options given to me, were 3 fractions and 0 and -9. :(

Answer 11

x value I mean...

Answer 12

wait is the function\[f(x)=\frac{1}{3}x^3-x^2-3x~~?\]

Answer 13

i substituted them, so the lesser results of either -1 or 2 is my minimum value?

Answer 14

yes that is the function.

Answer 15

because the 1st derivative should be \[f'(x)=x^2-2x-3\]

Answer 16

oh. my bad.

Answer 17

so x=3 and x=-1 right?

Answer 18

yup

Answer 19

yes it is...

Answer 20

so my result for x=3 is -9 and x=-1 is 5/3 so i pick 5/3 because it is smaller?

Answer 21

yup.. you're right...

Answer 22

oh wait, in terms of negativity, then -9 is smaller right?

Answer 23

take note your interval

Answer 24

x=3 is outside your interval

Answer 25

oh! Ok, so the answer must be within the interval. of course. Thank you!

Answer 26

yup... np