Question

Calculus help?

Given the derivative of f(x),
f ‘(x) = x^3 + 2x^2 – 8x – 8,
a) Use your calculator to determine one x value where f(x) has a relative extrema. Your value should be correct to 3 decimal places.

b) Classify the x value as a relative maximum or minimum.
Justify your answer in a complete sentence.

why do they tell me to use a calculator? am I supposed to insert a number to make it equal 0??

Answer 1

Relative extrema are found when f'(x)=0

Answer 2

Then, if f''(x) is negative, it's a local maximum. If f''(x) is positive, it's a local minimum.

Answer 3

P.S. If f''(x) is zero, double-check your calculation xD

Answer 4

so I need to factor it right?

Answer 5

I'm not sure, because normal schools won't ask you to factorize a cubic equation...

Answer 6

Well, it says "use a calculator", so I'm not sure how to do it.

Answer 7

0=x^2(x+2) 8(x-1)

is this correct?

Answer 8

yeah thats whats confusing me?

Answer 9

@ganeshie8 @dumbcow

ADMIN: @hartnn @satellite73 @.Sam. @amistre64 @Callisto

Answer 10

Please ban meghen_elodie.

Answer 11

You can not factor it easily, you may check the graph of this function on https://www.desmos.com/calculator. In this graph, I see we have 3 points of x, one of them maybe 9/4. Other points you should estimate or solve by yourself. Hint: (x-9/4)(Ax^2+Bx+C), then find A,B,C easily

Answer 12

(Meghen: you're lucky, no admins are on now, but I've reported you, so don't worry)

Answer 13

I am not sure how @linh412986 , kc can you see the points?

Answer 14

@linh412986 adding a full-stop has made the website unaccessable, and the x-intercept is definitely not 9/4

Answer 15

https://www.desmos.com/calculator/rqnvfbejsk

Answer 16

@VIbarguen1 where did you get this question? :)

Answer 17

@VIbarguen1 The purpose of question a) is explained by @kc_kennylau , we must find the solution of f' = 0 to find the extrema points. In order to solve for f' = 0, we solve the cubic equation. But it is hard to solve it, instead of it, we estimate the solution of f'=0 by graphing on desmos.
@kc_kennylau I made a mistake, one solution should be 5/2

Answer 18

@linh412986 neither is it 5/2.

Answer 19

it's homework :(

Answer 20

@linh412986 that is just approximation. the zeros calculated by WolframAlpha are here:

http://www.wolframalpha.com/input/?i=factorize+x%5E3%2B2x%5E2-8x-8

Answer 21

there are 2 ways to use calculator to find out the roots

Answer 22

Yes, that's just an approximation from graph. But the point here is that requirement of using calculator to solve :)

Answer 23

so kc you mean 2.494, -0.890, and 3.603 ?

Answer 24

https://en.wikipedia.org/wiki/Newton's_method

Answer 25

lin, I don't know how

Answer 26

@Loser66 . can you teach how to use calculator for this case?

Answer 27

it doesn't show it right

Answer 28

1/ hit the key Y= ..... then type the expression and then hit the trace key to find the roots
2/use program key to solve the cubic

Answer 29

You can use the Newton's method:
\[\Huge x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\](where \(x_{n+1}\) is the more accurate version of \(x_n\))

https://en.wikipedia.org/wiki/Newton's_method

Answer 30

you can hit the table key to see the table of the graph then up or down to see where y =0

Answer 31

by using program, I have x1 = -3.603875472
x2= -0.8900837358
x3= 2.493959270

Answer 32

I can do the same on an online cal, like the one lin put up ys?

Answer 33

Is the Newton's Method allowed?

Answer 34

thats what kc sent me with wolframalpha

Answer 35

I don't know. @Loser66 is it?

Answer 36

I don't follow, sorry friends, just support you in finding the roots. Follow @kc_kennylau guiding, please

Answer 37

What has your school taught you? Maybe you can look from the previous examples in your book?

Answer 38

There isn't any example quite like this one in the book. and I do online schooling so the book is the only thing I have to follow :(

Answer 39

at least, I can get this problem done.

Answer 40

trust me, it's been thanks to the help I've gotten on here that can I understand how to do half of the problems but some I still can't solve.

Answer 41

http://www.youtube.com/watch?v=bITQg_k7K0c

Answer 42

thanks you guys, for all the help

Answer 43

no problem at all :)

Answer 44

http://www.youtube.com/watch?v=8MfMiLbunuY
if you are allowed to use calculator to solve quadratic and cubic, you should have those program on yours.
for your problem f"(x) = 0 when x = 1.097167541
and x = -2.430500874
so far, you have 5 candidates for critical points ( extrema and local max/min)
to find value of x , you just take integral of f'(x) to find out the original function of f (x)
then plug those values to find the values of f(x). Compare them to consider which is extreme max/min, which is local one . DAT SIT

Answer 45

I did that and I got 2.494 to work. Am I right?

Answer 46

\[\int x^3 +2x^2-8x-8 dx = \frac{x^4}{4}+\frac{2x^3}{3}-4x^2-8x +C\]

Answer 47

ok got it, thank you!!

Answer 48

for x = -3.604 f(-3.604) = -12.154
do the same with the leftover.

Answer 49

ok I will

Answer 50

good luck

Answer 51

thanks