Graph f'(x) = 4x^2 (x+3): Troubleshooting Logic

[jwxie]

Given equation
f(x) = x^4 + 4x^3 - 2

so f ' (x) = 4x^3 + 12x^2
which simplified to
f ' (x) = 4x^2 (x+3), thus our critical points are 0 and -3

I will use this to sketch the curve
f ' (x) = 4x^2 (x+3)

Results are below.
According to my "sketch", the sequence is - , +, +, +
B means "before"
A means "after"
I used a very close number to generate this list

But I could not generate any sketch because (0,0) pair cannot have a +, +, + sequence. There must be a - sign, meaning the correct sequence should be -, +, -, +.
I need to hite (0,0) but there is no way as (-3,0) is increasing.

My calculator shows a different sequence. What is wrong with my sketch logic here?

 

[HallsofIvy]

Given equation
f(x) = x^4 + 4x^3 - 2

so f ' (x) = 4x^3 + 12x^2
which simplified to
f ' (x) = 4x^2 (x+3), thus our critical points are 0 and -3

I will use this to sketch the curve
f ' (x) = 4x^2 (x+3)

Don't you mean the original curve, y= x^4- 4x^2- 2?

Results are below.
According to my "sketch", the sequence is - , +, +, +
B means "before"
A means "after"
I used a very close number to generate this list

But I could not generate any sketch because (0,0) pair cannot have a +, +, + sequence. There must be a - sign, meaning the correct sequence should be -, +, -, +.
I need to hite (0,0) but there is no way as (-3,0) is increasing.

My calculator shows a different sequence. What is wrong with my sketch logic here?

If you actually are graphing y= 4x^2(x+3), then, yes, the graph is below the x-axis for x< -3 and above the x-axis for x> -3. The graph is then tangent to the x-axis at x= 0. It drops down from a maximum value (at (-2, 16)) to (0,0) but does NOT cross the axis there. It goes back up again.

 

[jwxie]

but as you see, in order to test the +, -, i did number line test which used the first derivative.

 

[jwxie]

i just found the problem.
the test was corrected but the y value was wrong...