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

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In summary, the equation f(x) = x^4 + 4x^3 - 2 has a first derivative of f ' (x) = 4x^3 + 12x^2, which simplified to f ' (x) = 4x^2 (x+3). The critical points are 0 and -3. When using this to sketch the curve, the sequence is -, +, +, +, and the graph is below the x-axis for x< -3 and above the x-axis for x> -3. However, in order to test the +, -, i did number line test which used the first derivative. The test was corrected, but the y value was wrong.
  • #1
jwxie
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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?
 

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  • #2
jwxie said:
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.
 
  • #3
but as you see, in order to test the +, -, i did number line test which used the first derivative.
 
  • #4
i just found the problem.
the test was corrected but the y value was wrong...
 

1. What does the function f'(x) represent?

The function f'(x) represents the derivative of the original function f(x). It measures the rate of change of the function at any given point.

2. Why is the derivative of f(x) important in this equation?

The derivative is important because it helps us understand the behavior of the original function. In this case, the derivative tells us the slope of the tangent line at any point on the graph of f(x).

3. How do I find the derivative of f(x) = 4x^2 (x+3)?

To find the derivative, you can use the power rule, which states that for a function of the form f(x) = ax^n, the derivative is f'(x) = anx^(n-1). In this case, the derivative of f(x) = 4x^2 (x+3) would be f'(x) = 8x^2 + 12x.

4. What does it mean if f'(x) = 0 in this equation?

If f'(x) = 0, it means that the slope of the tangent line at that point is equal to 0. This could indicate a point of maximum or minimum on the graph of f(x).

5. How can I use the derivative to troubleshoot any issues with the logic of this graph?

If you notice any issues with the logic of the graph, such as discontinuities or unexpected behavior, you can analyze the derivative to see where the function may be changing rapidly or approaching a maximum or minimum. This can help identify any errors in the logic of the graph.

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