If f(x) = 3x2 + 2x3 , then which of the following statements is true?

Question

f(x) has a local minimum at (0, 0) and a local minimum at (-1, 1)
f(x) has a local minimum at (0, 0) and a local maximum at (-1, 1)
f(x) has a local minimum at (0, 0) and a local maximum at (1, 1)
f(x) has a local maximum at (0, 0) and a local maximum at (-1, 1)

This is not my homework
I'm not even in enrolled in a Calculus class.

I'm tring to teach myself this summer and pass the class next year.

f(x) = 3x^2 + 2x^3

sorry about that

Answer 1

f(x) = 3x^2+2x^3
df/dx = 6x + 6x^2
6(-1)+6((-1)^2) = 0
6(0)+6(0^2) = 0

f(0) = 0
f(-1) = 3((-1)^2)+2((-1)^3) = 3 - 2 = 1

f(-1.1) = 3.3 + -1.331 = 0.638
f(-0.9) = 2.43 + -1.458 = 0.972
f(-0.1) = 0.03 + -0.002 = 0.028
f(0.1) = = 0.03 + 0.002 = 0.032

I'm a bit lazy to do the whole analysis, but I can safely assure you that f(x) has a local minimum at (0, 0) and a local maximum at (-1, 1).

Answer 2

The second statement is true.

f'(x)=6x+6x^2
Factor it and get f(x)=6x(1+x). To have a maximum/minimum, the derivative must equal zero. Hence, 6x(x+1)=0 at x=0 and x=-1. To determine if it is a maximum or a minimum, I believe you look at the second derivative: f''(x)=6+12x. When x=0, f">0 hence a local max. When x=-1, f"(-1)=-6<0, hence a local min.

You can also graph the equation and look at it from there. Have fun.

Answer 3

f(x) = 3x^2 + 2x^3
f '(x) = 6x + 6x^2 = 6x(1 + x)
f '(x) = 0 if x=0 or x=-1
If x=0 then f(x) = f(0) = 0 (Gives (0 , 0).
If x=-1 then f(x) = f(-1) = 3*(-1)^2 + 2*(-1)^3 = 3-2=1.
It gives point (-1,1)
When -1 f(x) has a local minimum at (0, 0) and a local maximum at (-1, 1)

Answer 4

You need to look at the second derivative to tell if a point is a max, min or point of inflection.

f'(x) = 6x + 6x^2 = 6(x)(1+x) so it has zeros at -1 and 0.

Check f''(x) = 6 + 12x
f''(-1) = -6 < 0 so the graph is concave down a x = -1. That makes -1 a relative max.
f''(0) = 6 > 0 so the graph is concave up at x = 0. That makes x and relative max.

Answer 5

Well I've been to some functions where the local minimum is 2.

Since you're asking this question on a sunday night, I think it's safe to say that by tomorrow morning you're in deep doodoo for not finishing your homework!

Answer 6

Since the function has no variables, it is an integer, single point, function with no minimum or maximum. All options presented appear to be invalid.

Answer 7

If 10 are pretend a million is authentic (#10) making 9 pretend If 9 are pretend 2 are authentic. (# 10 and 9) making 8 pretend if 8 are pretend 3 are authentic. (#10, 9, and eight) making 7 pretend it is going down till 5 5 are pretend 5 are authentic making 5 pretend solutions a million-5 are authentic

Answer 8

"f(x) = 3x2 + 2x3"
If this is what you mean by a function, then I'm really sorry because it is not a function in the first place and it equals just "6 + 6" = "12"
If you mean something else, then be clear in your function expression.

Answer 9

to get maximum & minimum value you need to differentiate thefunction

Given f(x) = 3x^2 + 2x^3
dy/dx = 3.2.x + 2.3.x^2
or dy/dx = 6x+6x^2

now evaluate dy/dx for (0,0), (1,1) and (-1,1)

and the values are dy/dx (0,0) = 0
dy/dx (1,1) = 12
dy/dx (-1,1) = -6+6 = 0

so third choice is correct.. i.e.

f(x( has a local minimum at (0,0) and a local maximum at (1,1)

Hope this comes to your help

Answer 10

third choice (0,0) (1,1)