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.

To find the critical numbers of the function f(x) = e^(-0.5x^2), we need to find the values of x where the derivative of the function is equal to zero or does not exist. The critical numbers represent the potential locations of local extrema.

(a) Finding the critical numbers:
1. Take the derivative of f(x) with respect to x:
f'(x) = (-0.5x^2)' * e^(-0.5x^2)
= -x * e^(-0.5x^2)

2. Set the derivative equal to zero and solve for x:
-x * e^(-0.5x^2) = 0

Since the exponential function e^(-0.5x^2) is always positive, the only way for the product to be zero is if x = 0.

Therefore, the critical number is x = 0.

(b) Determining where f is increasing or decreasing:
To determine where f is increasing or decreasing, we need to analyze the sign of the derivative.

1. Choose a test point to the left of x = 0, such as x = -1, and substitute it into the derivative:
f'(-1) = -(-1) * e^(-0.5(-1)^2) = -e^(-0.5)

Since e^(-0.5) is positive, f'(-1) is negative. This means that f(x) is decreasing to the left of x = 0.

2. Choose a test point to the right of x = 0, such as x = 1, and substitute it into the derivative:
f'(1) = -(1) * e^(-0.5(1)^2) = -e^(-0.5)

Again, since e^(-0.5) is positive, f'(1) is negative. This means that f(x) is also decreasing to the right of x = 0.

Therefore, f(x) is decreasing on the entire interval from negative infinity to positive infinity.

(c) Finding local extrema:
Since f(x) is decreasing everywhere, it does not have any local maxima or minima.

(d) Determining concavity and inflection points:
To determine where f is concave up or concave down, we need to analyze the sign of the second derivative.

1. Take the derivative of f'(x):
f''(x) = (-x * e^(-0.5x^2))'

Applying the product rule, we get:
f''(x) = -x * (-0.5x^2)' * e^(-0.5x^2) + (-x) * e^(-0.5x^2) * (-0.5x^2)'

Simplifying, we have:
f''(x) = x^2 * e^(-0.5x^2) - 0.5 * e^(-0.5x^2)

2. Set the second derivative equal to zero and solve for x:
x^2 * e^(-0.5x^2) - 0.5 * e^(-0.5x^2) = 0

Factor out e^(-0.5x^2):
e^(-0.5x^2) * (x^2 - 0.5) = 0

Either e^(-0.5x^2) = 0, which is not possible, or (x^2 - 0.5) = 0.

Solving x^2 - 0.5 = 0, we get x = ±√(0.5).

Therefore, the potential inflection points are x = ±√(0.5).

(d) Determining asymptotes:
Since the exponential function e^(-0.5x^2) does not tend to infinity or negative infinity as x approaches any value, there are no horizontal asymptotes.

Also, since the function does not have any vertical jumps or breaks, there are no vertical asymptotes.

Therefore, there are no horizontal or vertical asymptotes for f(x) = e^(-0.5x^2).

Please let me know if you need further clarification or assistance.

To find the critical numbers of the function f(x), we first need to find its derivative f'(x) and determine where it equals zero or is undefined.

Given the function f(x) = e^(-0.5x^2), we can find its derivative as follows:

f'(x) = d/dx (e^(-0.5x^2)).

Using the chain rule, we differentiate the exponential function and the power function:

f'(x) = e^(-0.5x^2) * d/dx (-0.5x^2).

Simplifying further:

f'(x) = e^(-0.5x^2) * (-x).

To find the critical numbers, we set f'(x) = 0 and solve for x:

0 = e^(-0.5x^2) * (-x).

Since e^(-0.5x^2) is always positive for real values of x, the equation simplifies to:

0 = -x.

From this equation, we can see that x = 0 is the only critical number.

Now, let's determine whether the function is increasing or decreasing on various intervals. To do this, we need to examine the sign of the derivative f'(x) in those intervals.

For x < 0, the derivative is negative (because of the negative sign in front of x in f'(x)), indicating that the function is decreasing in this interval.

For x > 0, the derivative is positive, indicating that the function is increasing in this interval.

To find the local extrema, we need to examine the behavior of the function f(x) around the critical number x = 0. We can do this by looking at the intervals on either side of x = 0.

For x < 0, the function is decreasing.

For x > 0, the function is increasing.

Therefore, x = 0 does not correspond to a local maximum or minimum.

Moving on to the concavity of the function, we need to find the second derivative f''(x). Let's compute it:

f''(x) = d^2/dx^2 (e^(-0.5x^2)).

Using the chain rule again, we differentiate the exponential function and the power function:

f''(x) = e^(-0.5x^2) * d^2/dx^2 (-0.5x^2).

Simplifying further:

f''(x) = e^(-0.5x^2) * (0.5 - x^2).

To find where the function is concave up or down, we set f''(x) = 0 and solve for x:

0 = e^(-0.5x^2) * (0.5 - x^2).

Since e^(-0.5x^2) is always positive for real values of x, the equation simplifies to:

0.5 - x^2 = 0.

Rearranging the equation, we find:

x^2 = 0.5.

Taking the square root of both sides, we obtain:

x = ±√(0.5).

So, the inflection points occur at x = √(0.5) and x = -√(0.5).

Now let's determine the concavity of the function in different intervals.

For x < -√(0.5), the second derivative f''(x) is positive, indicating that the function is concave up in this interval.

For -√(0.5) < x < √(0.5), the second derivative f''(x) is negative, indicating that the function is concave down in this interval.

For x > √(0.5), the second derivative f''(x) is positive again, indicating that the function is concave up in this interval.

Regarding the asymptotes of the function:

The function f(x) = e^(-0.5x^2) does not have any horizontal asymptotes because as x approaches positive or negative infinity, the exponential term tends to zero, making the function approach zero as well.

The function f(x) = e^(-0.5x^2) does not have any vertical asymptotes because the exponential term is always positive, and there are no values of x that would make the denominator of the function equal to zero.

In summary:

(a) The critical number of f(x) = e^(-0.5x^2) is x = 0.
The function is increasing on the interval (0, ∞).
The function is decreasing on the interval (-∞, 0).
There are no local maxima or minima.

(b) The function is concave up on the interval (-∞, -√(0.5)) ∪ (√(0.5), ∞).
The function is concave down on the interval (-√(0.5), √(0.5)).
The inflection points are x = √(0.5) and x = -√(0.5).

The horizontal asymptotes are y = 0.
The function does not have any vertical asymptotes.

I hope this explanation helps clarify the solutions to the problem. Let me know if you have any further questions!