how do I graph this using the first and second derivative and three sign charts.
i have to label the inflection point, extreme values and the intercepts also.
f(x)= x^3-6x^2
slope = 3 x^2 -12 x =3 x (x-4)
second derivative (curvature)
= 6 x - 12
so the slope is 0 at x = 0 and at x = 4
at x = 0
is that a min or a max or in inflection point?
second derivative = -12
That is a maximum, starts down from there
at x = 4 is that a min or a max or an inflection point?
second derivative = +12
so that is a minimum, starts up fro there.
Oh, intercepts
y = x^3-6x^2
0 = x^2 (x-6)
bounced off x axis at x = 0
crossed x axis at x = 6
To graph the function f(x) = x^3 - 6x^2 using the first and second derivative as well as three sign charts, you can follow these steps:
Step 1: Find the first derivative f'(x) by differentiating f(x) with respect to x:
f'(x) = 3x^2 - 12x
Step 2: Find the critical points by setting f'(x) equal to zero and solving for x:
3x^2 - 12x = 0
3x(x - 4) = 0
x = 0 or x = 4
Step 3: Create a number line and mark the critical points as well as additional test points in each of the intervals they create.
- Choose a test value less than 0, like x = -1.
- Choose a test value between 0 and 4, like x = 2.
- Choose a test value greater than 4, like x = 5.
Step 4: Determine the sign of f'(x) in each interval by plugging in the test values.
In the interval (-∞, 0):
- Substitute x = -1 into f'(x): f'(-1) = 3(-1)^2 - 12(-1) = 3 + 12 = 15 (positive)
- Therefore, the sign of f'(x) is positive in this interval.
In the interval (0, 4):
- Substitute x = 2 into f'(x): f'(2) = 3(2)^2 - 12(2) = 12 - 24 = -12 (negative)
- Therefore, the sign of f'(x) is negative in this interval.
In the interval (4, ∞):
- Substitute x = 5 into f'(x): f'(5) = 3(5)^2 - 12(5) = 75 - 60 = 15 (positive)
- Therefore, the sign of f'(x) is positive in this interval.
Step 5: Create a sign chart for f'(x) using the signs determined in step 4.
| (-∞, 0) | (0, 4) | (4, ∞) |
--------------------------------
f'| + | - | + |
Step 6: Find the second derivative f''(x) by differentiating f'(x) with respect to x:
f''(x) = 6x - 12
Step 7: Determine the sign of f''(x) in each interval by plugging in the test values.
In the interval (-∞, 0):
- Substitute x = -1 into f''(x): f''(-1) = 6(-1) - 12 = -6 - 12 = -18 (negative)
- Therefore, the sign of f''(x) is negative in this interval.
In the interval (0, 4):
- Substitute x = 2 into f''(x): f''(2) = 6(2) - 12 = 12 - 12 = 0 (zero)
- Therefore, the sign of f''(x) is zero in this interval.
In the interval (4, ∞):
- Substitute x = 5 into f''(x): f''(5) = 6(5) - 12 = 30 - 12 = 18 (positive)
- Therefore, the sign of f''(x) is positive in this interval.
Step 8: Create a sign chart for f''(x) using the signs determined in step 7.
| (-∞, 0) | (0, 4) | (4, ∞) |
--------------------------------
f''| - | 0 | + |
Step 9: Using the sign charts, identify the intervals where f(x) is increasing or decreasing and where it is concave up or concave down:
- f(x) is increasing in the interval (0, 4) because f'(x) is negative.
- f(x) is decreasing in the interval (-∞, 0) and (4, ∞) because f'(x) is positive.
- f(x) is concave up in the interval (4, ∞) because f''(x) is positive.
- f(x) is concave down in the interval (-∞, 0) because f''(x) is negative.
Step 10: Find the x-intercepts by solving f(x) = 0:
x^3 - 6x^2 = 0
x^2(x - 6) = 0
x = 0 or x = 6
Step 11: Find the y-intercept by plugging in x = 0 into f(x):
f(0) = (0)^3 - 6(0)^2 = 0
So, the y-intercept is (0, 0).
Step 12: Find the inflection point by solving f''(x) = 0:
6x - 12 = 0
6x = 12
x = 2
Substitute x = 2 into f(x) to find the corresponding y-value:
f(2) = (2)^3 - 6(2)^2 = 8 - 24 = -16
So, the inflection point is (2, -16).
Step 13: Plot the x-intercepts, y-intercept, and the inflection point on the graph.
Step 14: Finally, based on the sign charts and the intervals identified in step 9, sketch the graph of f(x) accordingly, considering the increasing/decreasing and concave up/down intervals.