Question

the absolute global maximum and absolute function value of x^3 - 3x^2 + 1 on the interval (-.5,4)

Answer 1

https://www.google.co.in/search?sourceid=chrome&ie=UTF-8&q=the+absolute+global+maximum+and+absolute+function+value+of+x%5E3+-+3x%5E2+%2B+1+on+the+interval+(-.5%2C4)

Answer 2

what are you asking?

Answer 3

please clarify

Answer 4

well im given these options

a) 225,49/16
b)2, 0
c)17,-3
d)1,-3
e) 17, 1/8

Answer 5

i need to find the absolute min and max values of the function on that interval

Answer 6

first take the derivative of the function

Answer 7

3(x-2)x

Answer 8

f(x) = x^3 - 3x^2 + 1
f'(x) = 3x^(2) - 6x

Answer 9

so we are looking on the interval
(-.5,4)

Step 1 - First check the domain of the derivative
what is it?

Answer 10

The domain is obviously
\[\mathbb{R}\]
so we dont have any critical points there

Step 2 - Set f'(x) = 0
what values of x make the equation = 0

Answer 11

its b

Answer 12

2 and zero

Answer 13

ok annonymous, can you just give me a quick rundown how

Answer 14

0 = 3(x-2)x
we notice
x = 0
x = 2
make the equation equal to 0
Are both numbers in the interval:
(-.5,4)

Yes they are so we will use both numbers to define the function in the interval

So remember

f'(x) > 0 the function is increasing
f'(x) < 0 the function is decreasing

so test numbers that are in the intervals of (-0.5, 0), (0 2), and (2, 4) to see if the function is positive or negative.

Best to make a chart

Answer 15

first find x when f'(x)=0
f'(x)=3x^2-6x
3x^2-6x=0
3x(x-2)=0
so we'll have critical points at x=0,2, that also means that we have absolute extremes at those points which are exactly the absolute max and min.

Answer 16

ok that was nice thanks

Answer 17

so lets test interval
(-0.5, 0)

f'(-0.2) =3((-0.2)-2)(-0.2)
so
3(-2.2)(-0.2) = a negative number thus
f'(x) < 0 thus the function is decreasing