Question

ind the extreme values of the function and where they occur y = x^3 - 3x^2 + 4x - 4

Answer 1

Calculus? :)
So uh, differentiate, yes?
Power rule?

Or do you need to be using the formal definition for derivative at this point?

Answer 2

Where we getting stuck?

Answer 3

Yes ugh
I know i have to find the derivative, set equal to zero and find the x and y
But im stuck at finding the derivative
Im not sure if i can use imaginary numbers in calculus

Answer 4

power rule

Answer 5

How would i use the power rule?

Answer 6

x^3---->3x^(3-1)

Answer 7

y' = 3x^2 - 6x + 4

b^2 - 4 ac < 0

@magepker728 , the equation has no real roots

Answer 8

Graphing is the easy way out :(
I need to learn how to do it algebraically

Answer 9

Not too familiar with power rule just yet?
You bring the exponent down in front as a multiplier,
and decrease the exponent by 1.

\(\large\rm \frac{d}{dx}x^n=nx^{n-1}\)
So like, for your first term,
\(\large\rm \frac{d}{dx}x^3=3x^{3-1}\)

k? :o

Answer 10

Oh, i mistook power rule for the product rule. I know how to do power rule but i guess my real problem is that i do not know how to find the x value(s) after i find the derivative

Answer 11

@zepdrix , this is the derivative : y' = 3x^2 - 6x + 4

to find the extrema 3x^2 -6x + 4 = 0

b^2 - 4ac < 0

No real roots only complex ( a + bi )

He wanna know how to solve it while the roots aren't real., got his problem ?

Answer 12

@ zepdrix... 3x^2 - 6x + 4.... I tried to factor but it didnt work then i tried the quadratic formula and i got imaginary numbers

Answer 13

@trojanpoem how do i solve it if the roots aren' t? And are imaginary numbers allowed in calulus ab?

Answer 14

Ok so no critical/extreme points :)
cool, seems like you figured it out.

Are inflection points considered extreme points?
I can't remember how they categorize this stuff...
Have you gone over second derivative stuff yet?

Answer 15

@1041904 , No, Complex roots for extrema makes no sense ( at least to me)>

@Zepdrix, The second derivative is for convavity

Answer 16

yes, im aware :)
what does the category of "extreme points" include though?
I guess I don't recall.

Answer 17

First derivative as the slope is zero in maxima , minima (extrema)

Answer 18

What would i put for my response for this question?

Answer 19

@Michele_Laino

Answer 20

Can you help with this one?
Same prompt
Y = 2/sqrt((1-5)^2)

Answer 21

y = 2/sqrt((1-5)^2) ?

y = 2/sqrt(16) = 2/4 = 0.5

Answer 22

Sorry messed up
Y = 2/sqrt(1-5x^2)

Answer 23

What are you trying to do ?

Answer 24

extrema ?

Answer 25

Same as the first problem

Answer 26

Y' = 2 * -0.5 (1-5x^2)^(-3/2) * -10x
= 10x (1-5x^2)^(-3/2)

x = 0

1-5x^2 = 0

-5x^2 = -1

x^2 = 1/5

x = +- sqrt(5)/5



Do the rest.

Answer 27

How did you find this: Y = 2/sqrt((1-5)^2)

Answer 28

I found the derivative using the power rule.

You may also use

1/2 sqrt(1-5x^2) * the derivative of what's inside the root.

Answer 29

Why does the numerator move to the front?

Answer 30

?

Answer 31

Y = 2/sqrt(1 - 5x^2)
Y' = 2(-0.5).......
Why is the 2 from the numerator being multiplied

Answer 32

sqrt = 0.5
1/sqrt = -0.5

Answer 33

Could have used the quotient rule?

Answer 34

Yeah,

Answer 35

But using the power rule here will help you avoid some unnecessary simplification later.

Answer 36

first question:
we have no real solutions, so there are not critical points, and since the first derivative is always positive, we can conclude that the function:
\(y=x^3-3x^2+4x-4\) is always an increasing function

Answer 37

Thank you @Michele_Laino @TrojanPoem, after i get 10x/(1-5x^2)^3/2 what do i do from there?

Answer 38

There is critical point when either the derivative is undefined or = zero

Answer 39

so 10x = 0
, 1-5x^2 = 0

Answer 40

Oh, thank you

Answer 41

I got sqrt(1/5)

Answer 42

-+ ( when taking roots)

sqrt(1)/sqrt(5) = sqrt(5)/5 (convention) no roots in den

Answer 43

Why cant i square root it as one fraction?

Answer 44

I get -1 for both positive and negative values of sqrt(5)/5

2/1 - 1 = 0

(Sqrt(5)/5, 0) is my extreme value?

Answer 45

I'ts not one to one function like

this y = x^2

It won't be a problem until you're finding inverse

(-sqrt(5)/5, -1) , (sqrt(5)/5, -1) ( Ididn't calculate it took your results)