Suppose that f(x)=3x^3+3x. Find all critical values of f. Then use interval notation to state when f(x) is increasing and when f(x) is decreasing and to state when f(x) is concave up and concave down. Find the local maxima and local minima. Find all vertical and horizontal asymptotes as well as all inflection points.
f' = 9x^2+3
f" = 18x
max/min when f' = 0
increasing when f' > 0
concave up when f" > 0
Now go for it.
To find the critical values of a function, we need to find the values of x at which either the derivative is equal to zero or undefined.
1. Find the derivative of f(x):
f'(x) = 9x² + 3
2. Set f'(x) equal to zero and solve for x:
9x² + 3 = 0
9x² = -3
x² = -3/9
x² = -1/3
x = ± √(-1/3)
Since the square root of a negative number is not defined in the real number system, there are no critical values at which the derivative is equal to zero.
Now, let's determine the intervals where f(x) is increasing or decreasing, and where it is concave up or concave down.
3. Determine the sign of the derivative f'(x) for different intervals:
To find when f(x) is increasing or decreasing, we need to determine when f'(x) is positive or negative.
If x < -√(-1/3): Choose a value smaller than -√(-1/3) and substitute it into f'(x).
For example, let's choose x = -2.
f'(-2) = 9(-2)² + 3 = 36 + 3 = 39 (positive)
Therefore, f'(x) is positive for x < -√(-1/3), indicating that f(x) is increasing in this interval.
If x > -√(-1/3): Choose a value larger than -√(-1/3) and substitute it into f'(x).
For example, let's choose x = 2.
f'(2) = 9(2)² + 3 = 36 + 3 = 39 (positive)
Therefore, f'(x) is positive for x > -√(-1/3), indicating that f(x) is increasing in this interval as well.
4. Determine when f(x) is concave up or concave down:
To find when f(x) is concave up or concave down, we need to determine when the second derivative f''(x) is positive or negative.
The second derivative is f''(x) = 18x.
Since the second derivative is always positive (18x > 0 for all x), f(x) is concave up for all x in its domain.
Now, let's find the local maxima and local minima.
5. Find the critical points:
Since we found earlier that there are no critical values where f'(x) equals zero or is undefined, there are no local maxima or minima for this function.
6. Find any vertical and horizontal asymptotes:
To find vertical asymptotes, we need to determine any x-values where the function approaches infinity or negative infinity.
Since f(x) is a polynomial, it does not have any vertical asymptotes.
To find horizontal asymptotes, we need to determine the behavior of the function as x approaches positive or negative infinity.
As x approaches positive or negative infinity, the dominant term in f(x) is 3x³. Therefore, there is a horizontal asymptote at y = 3x³.
7. Find any inflection points:
Inflection points occur when the concavity of the curve changes. In this case, as we found earlier, f(x) is concave up for all x in its domain. Therefore, there are no inflection points.
In summary:
- Critical values: None
- Increasing intervals: (-∞, ∞)
- Decreasing intervals: None
- Concave up intervals: (-∞, ∞)
- Concave down intervals: None
- Local maxima: None
- Local minima: None
- Vertical asymptotes: None
- Horizontal asymptotes: y = 3x³
- Inflection points: None