Please answer the following questions about the function
f(x)=e^(-0.5x^2)
Instructions: If you are asked to find x- or y-values, enter either a number, a list of numbers separated by commas, or None if there aren't any solutions. Use interval notation if you are asked to find an interval or union of intervals, and enter { } if the interval is empty.
(a) Find the critical numbers of f, where it is increasing and decreasing, and its local extrema.
Critical numbers x=
Increasing on the interval
Decreasing on the interval
Local maxima x=
Local minima x=
(b) Find where f is concave up, concave down, and has inflection points.
Concave up on the interval=
Concave down on the interval =
Inflection points x=
Find any horizontal and vertical asymptotes of f.
Horizontal asymptotes y=
Vertical asymptotes x=
My answers were:
Critical numbers x= none
Increasing on the interval = didn't have anything
Decreasing on the interval =(-INF,-0.5)U(-0.5,0.5)U(0.5,INF)
Local maxima x= none
Local minima x= none
Concave up on the interval= (-0.5,0)U(0.5,INF)
Concave down on the interval = (-INF,-0.5)U(0,0.5)
Inflection points x= none
Horizontal asymptotes y= none
Vertical asymptotes x=none
Amost none of them is right, only the local minima and the vertical asymtote, please help, I'm not sure how to do this problem.
Let's go step-by-step through the problem to find the correct answers:
(a) To find the critical numbers of f, we need to find the values of x where the derivative of f is equal to zero or undefined.
To start, let's find the derivative of f(x). The derivative of e^(-0.5x^2) can be calculated using the chain rule, which gives us f'(x) = -x*e^(-0.5x^2).
Next, we set f'(x) = 0 to find the critical numbers. Thus, -x*e^(-0.5x^2) = 0. Since e^(-0.5x^2) is always positive, this equation is satisfied when x = 0.
Hence, the critical number is x = 0.
To determine where the function is increasing or decreasing, we can use the first derivative test. We check the sign of the derivative on either side of the critical number.
Let's evaluate f'(-1) and f'(1):
f'(-1) = -(-1) * e^(-0.5*(-1)^2) = -e^(-0.5) < 0
f'(1) = -(1) * e^(-0.5*1^2) = -e^(-0.5) < 0
Since the derivative is negative on both sides of x = 0, the function is decreasing on the entire number line.
As for local extrema, since the function is decreasing on the entire number line, it does not have any local maxima or minima.
To summarize:
Critical numbers x = 0
Increasing on the interval: None
Decreasing on the interval: (-∞, ∞)
Local maxima x: None
Local minima x: None
(b) To find where f is concave up or concave down, we need to find the second derivative of f and examine its sign.
Let's find the second derivative. The derivative of f'(x) = -x*e^(-0.5x^2) can be calculated using the product rule, which gives us f''(x) = (-1)e^(-0.5x^2) + (-x)(-2x)e^(-0.5x^2) = e^(-0.5x^2) - 2x^2e^(-0.5x^2).
To determine concavity, we need to analyze the sign of the second derivative, f''(x).
Let's evaluate f''(-1) and f''(1):
f''(-1) = e^(-0.5(-1)^2) - 2(-1)^2e^(-0.5(-1)^2) = 1 - 2e^(-0.5) > 0
f''(1) = e^(-0.5(1)^2) - 2(1)^2e^(-0.5(1)^2) = 1 - 2e^(-0.5) > 0
Since the second derivative is positive on both sides of the critical number x = 0, the function is concave up on the entire number line.
Regarding inflection points, since the second derivative is always positive, there are no inflection points.
To summarize:
Concave up on the interval: (-∞, ∞)
Concave down on the interval: None
Inflection points x: None
Finally, let's find the horizontal and vertical asymptotes of f:
Since e^(-0.5x^2) approaches 0 as x approaches ±∞, there are no horizontal asymptotes.
However, there are no vertical asymptotes in the function f(x) = e^(-0.5x^2). It does not have any vertical asymptotes.
To summarize:
Horizontal asymptotes y: None
Vertical asymptotes x: None
So, the correct answers are:
(a)
Critical numbers x = 0
Increasing on the interval: None
Decreasing on the interval: (-∞, ∞)
Local maxima x: None
Local minima x: None
(b)
Concave up on the interval: (-∞, ∞)
Concave down on the interval: None
Inflection points x: None
Horizontal asymptotes y: None
Vertical asymptotes x: None
To find the critical numbers of the function f(x)=e^(-0.5x^2) and determine where it is increasing and decreasing, we need to first find the derivative of f(x).
Let's find the derivative of f(x) using the chain rule. The derivative of e^u, where u is a function of x, is given by d(e^u)/dx = (du/dx) * e^u.
To find the derivative of f(x), we need to determine the derivative of -0.5x^2, which is -x, and multiply it by e^(-0.5x^2). Using the chain rule, we have:
f'(x) = (-x) * e^(-0.5x^2)
Now, set f'(x) equal to zero and solve for x to find the critical numbers:
(-x) * e^(-0.5x^2) = 0
Since e^(-0.5x^2) is never zero, the only way the product can be zero is if -x = 0. Therefore, the critical number is x = 0.
To determine where the function is increasing and decreasing, we need to analyze the sign of the derivative on different intervals. We can do this by considering the sign of f'(x) and the behavior of the function around the critical point x = 0.
For x < 0, we have (-x) < 0. Since e^(-0.5x^2) is always positive, f'(x) < 0 for x < 0. This indicates that the function f(x) is decreasing on the interval (-∞, 0).
For x > 0, we have (-x) > 0. Again, since e^(-0.5x^2) is always positive, f'(x) > 0 for x > 0. This indicates that the function f(x) is increasing on the interval (0, ∞).
Now let's find the local extrema by analyzing the behavior of f(x) around the critical point x = 0. We'll find where the function changes from decreasing to increasing (local minimum) and from increasing to decreasing (local maximum).
Around x = 0, the function f(x) goes from negative to positive values. Therefore, we can conclude that f(x) has a local minimum at x = 0.
So, the answers to part (a) are:
Critical numbers x = 0
Increasing on the interval (0, ∞)
Decreasing on the interval (-∞, 0)
Local maxima x = None
Local minima x = 0
Moving on to part (b), to determine where f(x) is concave up and concave down, we need to find the second derivative of f(x).
Let's find the second derivative of f(x) by finding the derivative of f'(x).
To do this, we'll differentiate f'(x) = (-x) * e^(-0.5x^2) using the product rule.
f''(x) = -e^(-0.5x^2) + (-x) * (-0.5 * 2x * e^(-0.5x^2))
= -e^(-0.5x^2) + x^2 * e^(-0.5x^2)
Now, to find where f(x) is concave up and concave down, we need to consider the sign of the second derivative, f''(x).
For x < 0, we have (-e^(-0.5x^2)) < 0 and x^2 > 0. Therefore, f''(x) < 0 for x < 0, indicating that f(x) is concave down on the interval (-∞, 0).
For x > 0, we have (-e^(-0.5x^2)) < 0, but x^2 is also negative, resulting in x^2 * (-e^(-0.5x^2)) > 0. Therefore, f''(x) > 0 for x > 0, indicating that f(x) is concave up on the interval (0, ∞).
Now, let's look for potential inflection points by finding where the concavity changes. We need to determine where f''(x) changes sign.
Since f''(x) = -e^(-0.5x^2) + x^2 * e^(-0.5x^2), it only changes sign when -e^(-0.5x^2) = x^2 * e^(-0.5x^2).
Rearranging this expression, we have -1 = x^2 for x ≠ 0. However, this equation has no real solutions.
Hence, in part (b), the answers are as follows:
Concave up on the interval = (0, ∞)
Concave down on the interval = (-∞, 0)
Inflection points x = None
Moving on to the last part, let's find any horizontal and vertical asymptotes.
As x approaches -∞ or ∞, e^(-0.5x^2) approaches 0. Therefore, the function f(x) approaches 0 as x approaches -∞ or ∞.
Hence, the horizontal asymptote is y = 0.
Regarding vertical asymptotes, there are none since the function does not have any restrictions or approaches infinity for any specific x value.
Therefore, the answers are:
Horizontal asymptotes y = 0
Vertical asymptotes x = None
I hope this explanation helps! Let me know if you have any further questions.