Question

Help, question below

Answer 1

The function f(x) = 2 x^3 - 30 x^2 + 96 x + 9 has one local minimum and one local maximum.
This function has a local minimum at x equals______ with value_______
and a local maximum at x_______ equals with value______

Answer 2

@ranga help

Answer 3

will be abck soon. answering another question now.

Answer 4

okay, and i got -96 for min and -6 for max, but it says it's wrong

Answer 5

What are the x values of your minimum and maximum?

Answer 6

-96 and -6, but idk if it's right

Answer 7

Not the f(x) value but I was asking for the x value that you got for the critical points.

Answer 8

the derivative for the function 6x^2 -60x+96, but i find it hard to find the critical numbers

Answer 9

Factor out 6 and throw it away.
6x^2 -60x+96 = 0
6(x^2 - 10x + 16) = 0
x^2 - 10x + 16 = 0
x^2 -8x - 2x + 16 = 0
x(x - 8) - 2(x - 8) = 0
(x-8)(x-2) = 0
x =2 or x = 8

Put these two values in f(x) and find f(2) and f(8)

Answer 10

so the 8 is min and 2 is max

Answer 11

so would the values be 0?

Answer 12

8 is the x value where minimum occurs. But that is not the minimum value of the function. For that you have to find f(8)

2 is the x value where maximum occurs. But that is not the maximum value of the function. For that you have to find f(2)

Answer 13

No. Don't put the x values of 2 and 8 in the derivative.
Put it in the original function f(x)

Answer 14

157, 1033

Answer 15

Nope.
f(x) = 2 x^3 - 30 x^2 + 96 x + 9

f(2) = 2(2^3) - 30(2^2) + 96(2) + 9 = ?

f(8) = 2(8^3) - 30(8^2) + 96(8) + 9 = ?

Answer 16

97, -119

Answer 17

Thanks

Answer 18

how do u find the critical # for a f(x) = 6x+4x^-1.. the negative one thing set me off track

Answer 19

97 and -119 are correct.

So the answer should be:
This function has a local minimum at x equals___2___ with value__97_____ and a local maximum at x__8_____ equals with value__-119____

Answer 20

Thanks

Answer 21

f(x) = 6x+4x^-1

What is the derivative. Use the power rule for x^-1

Answer 22

would be 6-4x^-2 or 6-4x^-1*(-2)?

Answer 23

It is 6-4x^-2

Answer 24

so how do i find the critical # after that, im just not good at finding critical #s?

Answer 25

Set the derivative to zero.
drop the x^-2 to denominator so it becomes 1/x^2
6 - 4/x^2 = 0
solve for x