Question

Please help! I will fan and medal

Answer 1

Find the absolute extrema of each function on the given interval.
f(x)=x^3-14x-5 on the interval [-1,8]

Answer 2

find the derivative first

Answer 3

oh shoot lol that was the derivative lol

Answer 4

post the original function

Answer 5

x^3-7x^2-5x+10 on [-1,8]

Answer 6

alright the derivative of that will be:
3x^2 - 14x - 5

Answer 7

a lot easier to factor than x^3-14x-5

Answer 8

yeah lol, my bad

Answer 9

so can you factor 3x^2 - 14x - 5

Answer 10

(3x+1)(x-5)

Answer 11

yep

Answer 12

now solve for x

Answer 13

x=-1/3 and x=5

Answer 14

correct, do you know what you just found?

Answer 15

Isn't it when the slope is equal to 0?

Answer 16

what you found is known as the critical points

Answer 17

oh ok

Answer 18

now are they in the interval [-1,8] ?

Answer 19

yes

Answer 20

yep so plug the following numbers into x^3-7x^2-5x+10 (original function):
-1/3, 5, -1, 8
Can you figure out what numbers I chose?

Answer 21

Well the two are the critical numbers, and other points were at the beginning and the end of the interval?

Answer 22

yep

Answer 23

plug them into the original function now

Answer 24

ok

Answer 25

tag me when you get it

Answer 26

ok

Answer 27

ok so is it x=8 is the max and x=5 is the min?

Answer 28

for -1 I got the value 7
for -1/3 I got 10.85
for 5 I got -65
and for 8 I got 34

Answer 29

the max would be 34

Answer 30

First: exactly what does "absolute extrema" mean? Note that "absolute extrema" relates to an interval.

Answer 31

too be exact, the absolute maximum is 34 and will occur at x = 8

Answer 32

In this "absolute extrema" problem, you need to find two different kinds of extrema:

1. Those where the first derivative = 0
2. Those extrema that occur at one endpoint or the other.

Have you done both?

Really, if you want feedback on your work, you'll need to share your calculations.

Answer 33

yes we found when the derivative was 0 but what do you mean at one endpoint or the other? We found the derivative to be 3x^2-14x-5 and factored to get (3x+1)(x-5) than equaled both of those to 0 to get x=-1/3 and x=5

Answer 34

suggestion: Make a list of the critical values occurring in the given interval. calculate the associated function values.

Also evaluate the function itself at both of the endpoints.

To find the "abs. max.," identify the greatest function value on that interval.

To find the "abs. mom.," identify the smallest function value on that interval.

@kayders1997?

Answer 35

Ok yes that makes sense

Answer 36

I am asking you to LOCATE the relative max and min that occur within the given interval. You must then evaluate the function at each such value.

Next, you need to evaluate the function at each of the 2 endpoints.

Next, rank your ordered pairs in increasing size by function value.
The largest is the "abs. max." on that interval.

and so on.

Answer 37

ok

Answer 38

so you will be looking at -1 -1/3 5 and 8 right?

Answer 39

Since I have not seen your actual work, I'm not responding "right" or "wrong" to your listing of x-values.

Again, you must evaluate the given function at each x-value determined, whether those values are end point x values or are inside the given interval.

I would like to see four points: (-1, f(-1)), etc., and then I'd like to see your ranking of those 4 points in ascending order by y-coordinate.

Answer 40

ok so the points (-1,7) (-1/3,10.85)(5,-65) and (8,34) so the absolute min would be (5,-65) and the absolute max would be (8,34)

Answer 41

yes...same answer as me lol

Answer 42

Looks right, only thing is for each critical point, you should test to see if the function is actually changing from increasing/decreasing or decreasing/increasing at each crit point by plugging in any x value inside the interval for f ' (x) and see if it is + or - value

Answer 43

It could just do one of these. like in y=x^3