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.
Critical number x=0
Increasing on interval (-Inf,0)
Decreasing on interval (0,Inf)
Local maxima x=0
Local minima x=none
Concave up on (-Inf,-1)U(1,Inf)
Concave down on (-1,1)
Inflection points x=-1,x=1
Horizontal asymptotes y=0
Vertical asymptotes x=none
To find the critical numbers of the function f(x) = e^(-0.5x^2), we need to find the points where the derivative of the function equals zero or does not exist.
Let's start by finding the derivative of f(x) with respect to x:
f'(x) = d/dx(e^(-0.5x^2))
= -x * e^(-0.5x^2)
To find the critical numbers, we need to solve the equation f'(x) = 0. Setting f'(x) equal to zero and solving for x, we have:
-x * e^(-0.5x^2) = 0
Since the exponential term e^(-0.5x^2) is never zero for any value of x, the equation -x * e^(-0.5x^2) = 0 is satisfied when either -x = 0 or -0.5x^2 = 0.
From -x = 0, we find x = 0. This is a critical number of f(x) = e^(-0.5x^2).
From -0.5x^2 = 0, we can solve for x to find x = 0. Again, this is a critical number.
So, the critical numbers of f(x) are x = 0.
To determine where the function is increasing and decreasing, we can analyze the sign of the derivative f'(x).
For x < 0, the value of f'(x) is positive when x > 0, the value of f'(x) is negative. This means that the function f(x) is increasing on the interval (-inf,0) and decreasing on the interval (0,inf).
The local extrema occur at the critical points, which in this case is x = 0. Since the function is both increasing and decreasing near x = 0, there are no local maxima or minima.
Moving on to the concavity of the function, we need to find where the second derivative f''(x) equals zero or does not exist.
To find f''(x), we differentiate f'(x):
f''(x) = d/dx(-x * e^(-0.5x^2))
= -e^(-0.5x^2) + x^2 * e^(-0.5x^2)
Setting f''(x) equal to zero and solving for x, we have:
-e^(-0.5x^2) + x^2 * e^(-0.5x^2) = 0
Rearranging the equation, we get:
e^(-0.5x^2)(x^2 - 1) = 0
Since e^(-0.5x^2) is never zero, the equation x^2 - 1 = 0 is satisfied when x = -1 or x = 1.
Therefore, the inflection points of f(x) are x = -1 and x = 1.
To determine the concavity of the function, we analyze the sign of f''(x).
For x < -1, f''(x) is negative. For -1 < x < 1, f''(x) is positive. And for x > 1, f''(x) is negative.
Thus, f(x) is concave up on the interval (-1, 1) and concave down on the intervals (-inf, -1) and (1, inf).
Regarding asymptotes, let's analyze the behavior of f(x) as x approaches positive and negative infinity.
As x approaches infinity, e^(-0.5x^2) approaches zero. Therefore, the horizontal asymptote is y = 0 as x approaches positive infinity.
As x approaches negative infinity, e^(-0.5x^2) also approaches zero. Hence, again, the horizontal asymptote is y = 0 as x approaches negative infinity.
There are no vertical asymptotes for the function f(x) = e^(-0.5x^2).
Summarizing the findings:
(a)
Critical numbers x = 0
Increasing on the interval (-inf, 0)
Decreasing on the interval (0, inf)
Local maxima x = None
Local minima x = None
(b)
Concave up on the interval (-1, 1)
Concave down on the intervals (-inf, -1) and (1, inf)
Inflection points x = -1, 1
Horizontal asymptotes y = 0
Vertical asymptotes x = None