Question

Consider the function below
f(x)= 8x^4-28x+9
Find the critical numbers of f (if any).

Find the open interval on which the function is decreasing.

Find the open interval on which the function is increasing

Apply the First Derivative Test to identify the relative extremum.

Answer 1

you should factorize your equation first of all. and then draw the graph. you will able say increasing and decreasing intervals.

Answer 2

i didnt know how to solve for f'(x)=4(8x^3-7)=0

Answer 3

are you familiar with calculus - derivatives?

Answer 4

yea im taking it

Answer 5

ok so can you find the derivative of this function?

Answer 6

give me few minutes, i will send you my solution

Answer 7

f'(x)=32x^3-28

Answer 8

ok
to find critical values equate this to zero and solve for x

Answer 9

4(8x^3-7)=0
This is where i got stuck. How do i solve the the cube root?

Answer 10

x^3 = 28/32

use your calculator

Answer 11

- its the same as cube root of 7/8

Answer 12

so it's 0.956 and -0.956?

Answer 13

not not negative because x^3 is positive

Answer 14

oh ok. so there's only 1 critical point?

Answer 15

yes
If you have a graphic calculator you''l see that this is a minimum
The question asks you to identify the realive extrema using first derivative test

Answer 16

You need to find the slope of the curve each side of the minimum point by finding the sign of f' at values just less than 0.956 and just above
I suggest you use 0.95 and 0.96

Answer 17

if we go from negative to postiuve then its a minimum
- positive to negative is a maximum.

Answer 18

the critical value of f is the value of f(x) when x = 0.956

Answer 19

i still dont understand how u find the relative extremum

Answer 20

ACtually we've done things the wrong way around here.
The first part is to draw the graph of f to answer the first 3 parts

Answer 21

Do you understand why f' = 0 at the extrema?

Answer 22

is it because the slope is 0?

Answer 23

Yes f' gives the value of the slope at any point on the graph and at the extrema the slope is horizontal

Answer 24

what's the next step?

Answer 25

to identify the nature of the extremium you do the first derivative test
f'(x) = 32x^3 - 28

at x = 0.95 f' = 32(0.95)^3 - 28 = -0.564 - showing a negative slope just before the extremium
at x = 0.96 f' = 32(0.96)^3 - 28

try working this one out If its positive it shows that the point is a minimum

Answer 26

0.311 a positive slope

Answer 27

right
that proves its a minimum

Answer 28

so the point is (0.965,0.311)?

Answer 29

no the point is (0.956, f(0.956))
plug x = 0.956 into the function 8x^4 - 28x + 9

this is the value of f at the extremium

Answer 30

you should get -11 point something

Answer 31

ok so the relative extremum is ((0.956, -11.086)?

Answer 32

yes

Answer 33

and the intervals are
decreasing (-INF, -11.086)

- i'll let you workout the interval for increasing f

Answer 34

shouldn't it be (INF, 0.956)?

Answer 35

Hmmm the values of x you mean
Yes I think you are right

Answer 36

so decreasing is (-INF,0.956)

Answer 37

yes my mistake

Answer 38

i thought it would be positive infinity

Answer 39

No the values are all less than 0.956 so it does back to negative infinty

Answer 40

the increasing interval will be (0.956, INF)

Answer 41

Note the round parentheses because at 0.956 the slope is horizontal

Answer 42

i submitted my work but i got the critical numbers wong :/

Answer 43

did they give you the numbers?