a. Find the critical points of f(x)=6x^4 e^-x.
b. Use the First Derivative Test to locate the local maximum and minimum values.
c. Determine the intervals on which f is concave up or concave down. Identify any inflection points.
a. Identify all the critical points. Select the correct choice below and fill in any answer boxes within your choice.
A. The function has critical points at x=___
To find the critical points, we need to find the values of x where the derivative of f(x) is equal to 0 or undefined.
a. Find the derivative of f(x):
f'(x) = 24x^3 e^(-x) - 6x^4 e^(-x)
Next, set f'(x) equal to 0 and solve for x:
24x^3 e^(-x) - 6x^4 e^(-x) = 0
Factor out e^(-x):
e^(-x)(24x^3 - 6x^4) = 0
Set each factor equal to 0:
e^(-x) = 0 (this is not possible)
24x^3 - 6x^4 = 0
Factor out 6x^3:
6x^3(4 - x) = 0
Set each factor equal to 0:
6x^3 = 0
x = 0 (this is a critical point)
4 - x = 0
x = 4 (this is a critical point)
Therefore, the function has critical points at x = 0 and x = 4.
a. The function has critical points at x = 0 and x = 4.
b. To use the First Derivative Test, we need to determine the sign of the derivative to the left and right of each critical point.
For x < 0, we can choose x = -1:
f'(-1) = 24(-1)^3 e^(-(-1)) - 6(-1)^4 e^(-(-1))
f'(-1) = 24e + 6e
f'(-1) = 30e (positive)
For 0 < x < 4, we can choose x = 1:
f'(1) = 24(1)^3 e^(-1) - 6(1)^4 e^(-1)
f'(1) = 24e^(-1) - 6e^(-1)
f'(1) = 18e^(-1) (positive)
For x > 4, we can choose x = 5:
f'(5) = 24(5)^3 e^(-5) - 6(5)^4 e^(-5)
f'(5) = 600e^(-5) (positive)
Using the First Derivative Test:
- At x = 0, the derivative changes from positive to negative, which indicates a local maximum.
- At x = 4, the derivative changes from negative to positive, which indicates a local minimum.
c. To determine the intervals of concavity and identify any inflection points, we need to find the second derivative of f(x).
f''(x) = 72x^2 e^(-x) + 24x^3 e^(-x) - 12x^3 e^(-x) - 6x^4 e^(-x)
Simplifying,
f''(x) = 72x^2 e^(-x) + 12x^3 e^(-x) - 6x^4 e^(-x)
To find where the concavity changes, we need to find where the second derivative is equal to 0 or undefined.
Set f''(x) equal to 0:
72x^2 e^(-x) + 12x^3 e^(-x) - 6x^4 e^(-x) = 0
e^(-x)(72x^2 + 12x^3 - 6x^4) = 0
Factor out e^(-x):
e^(-x)(72x^2 + 12x^3 - 6x^4) = 0
This equation has no solutions for e^(-x) = 0, so we need to solve for the expression inside the parenthesis:
72x^2 + 12x^3 - 6x^4 = 0
We can factor out x^2:
x^2(72 + 12x - 6x^2) = 0
Set each factor equal to 0:
x^2 = 0
x = 0 (this is the same critical point we found earlier)
72 + 12x - 6x^2 = 0
This equation can be factored:
6(12 + 2x - x^2) = 0
Set each factor equal to 0:
12 + 2x - x^2 = 0
Using the quadratic formula,
x = (-2 ± √(2^2 - 4(-1)(12))) / (2)
x = (-2 ± √(4 + 48)) / 2
x = (-2 ± √52) / 2
x = (-2 ± 2√13) / 2
x = -1 ± √13
Therefore, the function has an inflection point at x = -1 - √13 and x = -1 + √13.
c. The function is concave up on the intervals (-∞, -1 - √13) and (-1 + √13, ∞), and concave down on the interval (-1 - √13, -1 + √13). The function has inflection points at x = -1 - √13 and x = -1 + √13.