You are given that
f(x) =1+ax+ax2
f′(x) =−(x+2)ax3
f′′(x) =(2x+6)ax4
and the constant a>0.
Write down all intervals on which the function is increasing, decreasing, concave up or concave down.
(Enter using notation (a, b). Use a comma to separate multiple intervals; e.g. (a, b), (c, d). Type inf and -inf to denote ∞ and −∞ respectively.)
increasing:
- sin responder
decreasing:
- sin responder
concave up:
- sin responder
concave down:
- sin responder
Usted ha realizado 0 de 2 intentos de envío
F. (10) (4 puntos posibles)
Sketch the graph of f(x)=1+ax+ax2 (same function as above) on −∞<x<∞ showing the horizontal and vertical asymptotes. While the sketch need not be to scale, please make sure the x− coordinate of all critical point(s) and inflection point(s) are accurate.
To determine the intervals on which the function is increasing, decreasing, concave up, or concave down, we need to analyze the first and second derivatives of the function.
1. Increasing Intervals:
The function is increasing when the first derivative, f'(x), is positive. Let's find the critical points by setting f'(x) = 0 and solving for x:
-(x + 2)ax^3 = 0
This equation gives us two critical points: x = 0 and x = -2.
Now let's analyze the sign of f'(x) in different intervals:
- For x < -2: Since f'(x) = -(x + 2)ax^3, and a > 0, f'(x) will be negative in this interval.
- For -2 < x < 0: f'(x) will be positive because we have a negative coefficient (-) multiplied by a positive value (a) and (x + 2) is negative in this interval.
- For x > 0: f'(x) will be positive because both the coefficient (-) and (x + 2) will be negative in this interval.
Therefore, the function is increasing on the interval (0, ∞).
2. Decreasing Intervals:
The function is decreasing when the first derivative, f'(x), is negative. From the analysis of increasing intervals, we can conclude that f'(x) is negative on the interval (-∞, -2).
Therefore, the function is decreasing on the interval (-∞, -2).
3. Concave Up Intervals:
The function is concave up when the second derivative, f''(x), is positive. Let's find the critical points by setting f''(x) = 0 and solving for x:
(2x + 6)ax^4 = 0
This equation gives us one critical point: x = -3.
Now let's analyze the sign of f''(x) in different intervals:
- For x < -3: Since f''(x) = (2x + 6)ax^4, and a > 0, f''(x) will be negative in this interval.
- For x > -3: f''(x) will be positive because both the coefficient (2) and (x + 3) will be positive in this interval.
Therefore, the function is concave up on the interval (-∞, -3).
4. Concave Down Intervals:
The function is concave down when the second derivative, f''(x), is negative. From the analysis of concave up intervals, we can conclude that f''(x) is negative on the interval (-3, ∞).
Therefore, the function is concave down on the interval (-3, ∞).
Now let's move on to sketching the graph of f(x) = 1 + ax + ax^2 and identifying the horizontal and vertical asymptotes.
To sketch the graph:
1. Determine the y-intercept: Substitute x = 0 into the equation f(x) = 1 + ax + ax^2 to find the y-intercept. The y-intercept is (0, 1).
2. Determine the x-intercepts: Set f(x) = 0 and solve for x. The equation ax^2 + ax + 1 = 0 can be solved using the quadratic formula.
3. Plot the y-intercept and x-intercepts on the graph.
4. Analyze the behavior of the graph as x approaches infinity (positive) and negative infinity. This will help identify the horizontal asymptotes.
5. Determine the critical points and inflection points. We have already found the critical points as x = 0, x = -2, and the inflection point as x = -3.
Regarding the vertical asymptotes, since the function is a polynomial, there are no vertical asymptotes.
The final step is to plot the graph of f(x) with the analyzed information. Unfortunately, as an AI text-based bot, I can't provide a visual representation. However, following the steps above will help you accurately sketch the graph and identify the asymptotes.
To determine the intervals where the function is increasing or decreasing, we need to find the sign of the derivative, f'(x).
Since f'(x) = -(x+2)ax^3, the sign of f'(x) is determined by the sign of (x+2).
When (x+2) > 0, f'(x) is positive, indicating that the function is increasing. So, the function is increasing for x > -2.
When (x+2) < 0, f'(x) is negative, indicating that the function is decreasing. So, the function is decreasing for x < -2.
To determine the intervals where the function is concave up or concave down, we need to find the sign of the second derivative, f''(x).
Since f''(x) = (2x+6)ax^4, the sign of f''(x) is determined by the sign of (2x+6).
When (2x+6) > 0, f''(x) is positive, indicating that the function is concave up. So, the function is concave up for x > -3.
When (2x+6) < 0, f''(x) is negative, indicating that the function is concave down. So, the function is concave down for x < -3.
Therefore, the intervals can be summarized as follows:
Increasing: ( -2, ∞ )
Decreasing: ( -∞, -2 )
Concave Up: ( -3, ∞ )
Concave Down: ( -∞, -3 )
Please note that the intervals are given in the form (a, b), where a and b represent the interval's endpoints.