Any people who know basic calculus?

Question

I need to know what you get for this problem. Here it is....

What are the x-values of the relative minimum(s) of the line x^3 - 3x +1 with the domain being [-3,3]?

I got x=1 but the quiz my instructor gave didn't have that as an answer. What am I missing?

Answer 1

y = x^3 - 3x + 1
y' = 3x^2 -3 = 3(x-1)(x+1)
so x = -1 is a max and x = 1 is a min
maybe you had the question wrong, maybe instructor made a mistake.

Answer 2

First, we take a first derivative.

f'(x) = 3x^2 - 3

We set it equal to 0 to get the extreme.

3x^2 - 3 = 0
3(x^2 - 1) = 0
3(x+1)(x-1) = 0
x=1,-1

So, since our domain is bounded, we've got 4 options... either, x = -3, 3, 1, -1

so let's plug these number in to see which is the case...
f(-3) = -27 + 9 + 1 = -17
f(3) = 27 - 9 + 1 = 19
f(1) = 1 - 3 + 1 = -1
f(-1) = -1 - 3(-1) + 1 = 3

so... the relative minimum over this domain is at -3

*** Note ***
Key concept is that you've gotta check the bounds when you have a limited domain.

Answer 3

x^3 - 3x +1
dy/dx =3x^2-3
3x^2-3= 0
x= +/- 1
At x = -1 there is a relative maximum of 3
At x= 1, there is a relative minimum of -1

The only thing I see wrong is calling x^3-3x+1 a line. You should call it a function. The other point is that there can only be one local minimum and one local maximum so why the (s) after minimum?

Answer 4

a local minimum would be a point where F'(x) =0 and is increasing and then decreses.

So first we find F'(x) which would be 3x^2 -3

So F'(x) is going to equal zero at 1 & -1
3(1)^2 -3 = 0
3(-1)^2-3 = )

Now you need to check if it is going from positive to negative to see if it is really a maximum or a minimum.

The local maxium would be -1 in this case.

I'll give you some example points to show this
3(-2)^2-3
3(4) -3
=9
3(0)^2 -3
0-3
= -3

So as you can see F'(x) goes from positive to negative at x = -1.

Answer 5

OK.

f(x)= x^3 - 3x + 1

Take the first derivative:

3x^2 - 3 = f'(x)

Minima/maxima are where f'(x)=0

So, 3x^2 = 3.

Thus, x=1 or -1

x=1 means f(1)= 1 - 3 + 1 = -1

x=-1 means f(-1)= -1 +3 + 1= +3

Therefore, (1, -1) is the relative minimum.

Done.