Consider the function f(x)= (3/4)x^4-x^3-3x^2+6x
Find the relative extrema for f(x); be sure to label each as a maximum or minimum. You do not need to find function values; just find the x-values.
Determine the interval(s) where f(x) is increasing (if any) and the interval(s) where f(x) is decreasing (if any). Show why there are exactly two inflection points for this function. Note: You do NOT need to find the inflection points.
Any help on this would be much appreciated!
f = (3/4)x^4-x^3-3x^2+6x
f' = 3x^3-3x^2-6x+6
f" = 9x^2-6x-6
extrema where f'=0 and f"≠0
f increasing where f' > 0
there are two inflection points because f" has two real roots.
To find the relative extrema of the function f(x) = (3/4)x^4 - x^3 - 3x^2 + 6x, we need to find the critical points, where the derivative of the function is either zero or undefined.
1. Find the derivative of f(x):
f'(x) = 4(3/4)x^3 - 3x^2 - 6x + 6
= 3x^3 - 3x^2 - 6x + 6
2. Set f'(x) equal to zero and solve for x:
3x^3 - 3x^2 - 6x + 6 = 0
Unfortunately, there is no simple way to find the solutions to this equation, so we'll have to approximate the solutions using numerical methods or a graphing calculator. Let's assume the three x-values for these solutions are a, b, and c.
3. Determine the sign of f'(x) in the intervals:
Now, divide the number line into four intervals based on the values of a, b, and c. Choose a test point in each interval (not including a, b, or c) and evaluate the sign of f'(x) at those points.
Interval 1: (-∞, a)
Choose a test point, x = 0
Evaluate f'(0):
f'(0) = 3(0)^3 - 3(0)^2 - 6(0) + 6 = 6
Since f'(0) > 0, f(x) is increasing in this interval.
Interval 2: (a, b)
Choose a test point, x = (a+b)/2
Evaluate f'((a+b)/2):
f'((a+b)/2) = 3((a+b)/2)^3 - 3((a+b)/2)^2 - 6((a+b)/2) + 6
Let's assume f'((a+b)/2) > 0
In this case, f(x) is increasing in this interval.
Interval 3: (b, c)
Choose a test point, x = (b+c)/2
Evaluate f'((b+c)/2):
f'((b+c)/2) = 3((b+c)/2)^3 - 3((b+c)/2)^2 - 6((b+c)/2) + 6
Let's assume f'((b+c)/2) < 0
In this case, f(x) is decreasing in this interval.
Interval 4: (c, ∞)
Choose a test point, x = 1
Evaluate f'(1):
f'(1) = 3(1)^3 - 3(1)^2 - 6(1) + 6 = -3
Since f'(1) < 0, f(x) is decreasing in this interval.
4. Determine the relative extrema:
Based on the intervals where f'(x) changes sign, we can determine the relative extrema.
Maximum: The maximum occurs at the transition from increasing to decreasing, so it is between intervals 2 and 3 (at x = b).
Minimum: The minimum occurs at the transition from decreasing to increasing, so it is between intervals 3 and 4 (at x = c).
5. Determine the intervals of increasing and decreasing:
f(x) is increasing on the intervals (-∞, a) and (a, b).
f(x) is decreasing on the intervals (b, c) and (c, ∞).
6. Reasoning for two inflection points:
An inflection point occurs where the concavity of the function changes. The concavity of a function changes when its second derivative changes sign.
To find the inflection points, we need to find the solutions to f''(x) = 0, where f''(x) is the second derivative of f(x).
First derivative of f'(x):
f''(x) = 9x^2 - 6x - 6
To find the inflection points, we need to solve f''(x) = 0. Assuming the solutions are d and e.
Reasoning for two inflection points: Since f''(x) is a quadratic equation, it can have at most two solutions. Therefore, there are exactly two inflection points for this function.
To find the relative extrema of a function, we need to find the critical points where the derivative of the function equals zero or is undefined. The critical points can be local maximum or minimum points.
1. Find the derivative of the function:
f'(x) = (3/4) * 4x^3 - 3x^2 - 6x + 6
= 3x^3 - 3x^2 - 6x + 6
2. Set the derivative equal to zero and solve for x to find the critical points:
3x^3 - 3x^2 - 6x + 6 = 0
To solve this equation, you can either use a numerical method or factor it if possible. Let's factor it for simplicity:
3x^3 - 3x^2 - 6x + 6 = 0
3(x^3 - x^2 - 2x + 2) = 0
Now, we can recognize that x = 1 is a solution. We can use synthetic division to find the other two solutions:
1 | 1 -1 -2 2
| 1 0 -2
|_____________
1 0 -2 0
So, we have (x - 1)(x^2 - 2) = 0
This gives us three possible critical points: x = 1, x = √2, and x = -√2.
3. To determine whether these critical points are relative maxima or minima, we need to apply the second derivative test. Find the second derivative of the function:
f''(x) = 9x^2 - 6x - 6
4. Evaluate the second derivative at each critical point:
f''(1) = 9(1)^2 - 6(1) - 6 = 9 - 6 - 6 = -3
f''(√2) = 9(√2)^2 - 6(√2) - 6 = 9(2) - 6(√2) - 6 = 12 - 6√2 - 6 ≈ -2.5 - 6√2
f''(-√2) = 9(-√2)^2 - 6(-√2) - 6 = 9(2) + 6√2 - 6 ≈ 12 + 6√2 - 6 ≈ -2.5 + 6√2
5. Apply the second derivative test:
- If f''(x) < 0, the point x is a relative maximum.
- If f''(x) > 0, the point x is a relative minimum.
For f''(1) = -3, we have a relative maximum at x = 1.
For f''(√2) and f''(-√2), we have no clear conclusions since the second derivative is not straightforward.
6. To determine the intervals of increasing and decreasing, we can use the first derivative test:
- If f'(x) > 0, the function is increasing on that interval.
- If f'(x) < 0, the function is decreasing on that interval.
To identify the intervals, we can analyze the sign of f'(x) using a sign chart or testing specific intervals.
- Test f'(x) for x < -√2: f'(-3) = (-)(-) - (+) + (+) = -
- Test f'(x) for -√2 < x < 1: f'(-1) = (-)(+) - (+) + (+) = -
- Test f'(x) for x > 1: f'(2) = (+)(+) - (+) - (+) = -
From these tests, we can conclude that f(x) is decreasing on the interval (-∞, -√2) and increasing on the intervals (-√2, 1) and (1, ∞).
7. Lastly, notice that finding the inflection points was not required for this task. However, we can determine the number of inflection points by looking at the concavity of the graph. Since the second derivative test did not provide definitive results for the critical points √2 and -√2, we cannot determine the exact number of inflection points. We only know that there are at least two inflection points based on the possibility of a change in concavity.
In summary:
- The relative maximum is at x = 1.
- The function is decreasing on the interval (-∞, -√2) and increasing on the intervals (-√2, 1) and (1, ∞).
- There are exactly two inflection points according to the problem statement, but their exact locations are not determined without further analysis.