Find the x-values where f(x)=x^4-8x^3 attains a relative minimum, relative maximum, and inflection points, if any. Justify your answers.
My derivative test didn't seem right.
To find the x-values where the function f(x) = x^4 - 8x^3 attains a relative minimum, relative maximum, and inflection points, we need to analyze its first and second derivatives.
First, let's calculate the first derivative of f(x) to determine the critical points and stationary points:
f'(x) = 4x^3 - 24x^2
The critical points occur where f'(x) = 0 or where the derivative is undefined. To find these critical points, we set f'(x) = 0:
4x^3 - 24x^2 = 0
Factoring out common terms, we get:
4x^2(x - 6) = 0
This equation holds true if either:
1) 4x^2 = 0, which gives us x = 0
2) x - 6 = 0, which gives us x = 6
So, we have two critical points: x = 0 and x = 6.
Now, let's calculate the second derivative of f(x):
f''(x) = 12x^2 - 48x
To classify the critical points as relative minimum, relative maximum, or neither, we evaluate the second derivative at these points.
For x = 0:
f''(0) = 12(0)^2 - 48(0) = 0
For x = 6:
f''(6) = 12(6)^2 - 48(6) = 72(6 - 4) = 144 > 0
Based on the second derivative test:
- If f''(x) > 0, then f(x) has a relative minimum at x.
- If f''(x) < 0, then f(x) has a relative maximum at x.
- If f''(x) = 0 or f''(x) does not exist, the test is inconclusive.
In this case, f''(0) = 0, which means the test is inconclusive for x = 0. However, f''(6) > 0, so f(x) has a relative minimum at x = 6.
To determine if there are any inflection points, we analyze the critical points of f''(x). Since f''(0) = 0, we need to examine the behavior of f''(x) around x = 0. This can be done by checking the signs of f''(x) values on either side of x = 0:
For x < 0:
f''(-1) = 12(-1)^2 - 48(-1) = 12 + 48 = 60 > 0
For x > 0:
f''(1) = 12(1)^2 - 48(1) = 12 - 48 = -36 < 0
Since the sign changes from positive to negative, x = 0 is an inflection point.
In summary, the function f(x) = x^4 - 8x^3 has:
- A relative minimum at x = 6
- An inflection point at x = 0 (neither relative minimum nor maximum)
Note: If your derivative test didn't seem right, double-check your calculations and the signs of the derivatives. It's important to be accurate when calculating and interpreting derivative values.