let f(x)=x^3(e^-x)
Answer using calculus, use graphing calculator only to check work.
a) Find the local and global extrema of f.
b) Find the intervals where f is increasing/decreasing.
c) Find the inflection points of f.
d) Find the intervals where f is concave up/down.
e) Find any asymptotic behavior (use limits to justify your answer)
Here's my work. Not sure I did it right so if someone could check. I don't know how to do e).
f'(x) = x^2(e^-x)(-x+3)=0
critical values are x=0 and x=3.
Not sure how to use these values to find part a.
f"(x)= x(e^-x)(-6x+x^2+6)
x=0, x=4.7321, x=1.2679 (find these with quadratic equation). These are the inflection points.
Increasing for x<3, until it hits x=3 which I think is the maximum (not sure if it's local or global extrema).
Decreasing for x>3, hit a local minimum at x=0 ( I'm not sure on these parts)
Concave up (0,1.2679), (4.7321, infinity)
Concave down (-infinity,0), (1.2679,4.7321)
I don't know how to do part e.
To find the local and global extrema of the function f(x)=x^3(e^-x), we need to find the critical points and determine whether they correspond to a maximum or minimum.
a) To find the critical points, we set the derivative f'(x)=0:
f'(x) = x^2(e^(-x))(-x+3) = 0
The critical values are x=0 and x=3. To determine whether these are local maxima or minima, we can use the second derivative test or evaluate the function at these points.
b) To find the intervals where f is increasing or decreasing, we need to analyze the sign of the derivative f'(x) in different regions.
For x < 0, the term (-x+3) is negative. Since x^2 and e^(-x) are always positive, f'(x) is negative, meaning the function is decreasing in this interval.
For 0 < x < 3, both x^2 and (-x+3) are positive. With e^(-x) being positive as well, f'(x) is positive, indicating that the function is increasing in this interval.
For x > 3, the term (-x+3) becomes negative again, while x^2 and e^(-x) remain positive. Thus, f'(x) is negative, and the function is decreasing in this interval.
c) To find the inflection points of f, we need to find the x-values where the second derivative f''(x) changes sign or equals zero.
The second derivative is given by:
f''(x) = x(e^(-x))(-6x+x^2+6)
Setting f''(x)=0 and solving for x gives us three solutions: x=0, x=4.7321, and x=1.2679. These are the potential inflection points.
d) To find the intervals where f is concave up or concave down, we need to analyze the sign of the second derivative f''(x) in different regions.
For x < 0, all three terms in f''(x) are negative. Thus, f''(x) is negative, and the function is concave down in this interval.
For 0 < x < 1.2679, x(e^-x) is positive, while (-6x+x^2+6) is negative. Therefore, f''(x) is positive, meaning the function is concave up in this interval.
For 1.2679 < x < 4.7321, both x(e^-x) and (-6x+x^2+6) are positive, making f''(x) positive. Hence, the function is concave up in this interval.
For x > 4.7321, x(e^-x) is negative, while (-6x+x^2+6) is positive. So, f''(x) is negative, and the function is concave down in this interval.
e) To determine the asymptotic behavior of f, we can analyze the limits as x approaches positive or negative infinity.
As x approaches negative infinity, e^-x approaches infinity, making the function f(x) approach negative infinity.
As x approaches positive infinity, e^-x approaches 0, and the function f(x) approaches positive infinity.
Therefore, f(x) approaches negative infinity as x approaches negative infinity and approaches positive infinity as x approaches positive infinity, indicating no asymptotes.
To verify these results using a graphing calculator, you can graph the function f(x) and observe the behavior of the function over the specified intervals and at the critical points.
To find the extrema of the function f(x) = x^3(e^-x), we need to find the critical points by setting the derivative equal to zero.
a) To find the critical points of f(x), we calculate f'(x) = 0:
f'(x) = x^2(e^-x)(-x+3)
Setting this equal to zero, we have x^2(e^-x)(-x+3) = 0
To solve this equation, we have two factors: x^2 = 0 and (e^-x)(-x+3) = 0.
For x^2 = 0, the only critical point is x = 0.
For (e^-x)(-x+3) = 0, we need to solve for x. However, since e^-x is never equal to zero, we can ignore that factor. Solving -x + 3 = 0, we get x = 3. Therefore, we have two critical points: x = 0 and x = 3.
To determine if these points are local extrema or global extrema, we need further analysis. We can use the second derivative test or examine the graph to determine if they are local maxima or minima. A graphing calculator can be used to visualize the graph of f(x) and determine the nature of the extrema.
b) To find the intervals where f is increasing and decreasing, we need to evaluate the sign of the derivative f'(x) in different intervals.
From f'(x) = x^2(e^-x)(-x+3), we see that if x < 0, then x^2 > 0 and (e^-x) > 0. Therefore, f'(x) < 0.
Between 0 and 3, we know that (e^-x) > 0 since exponential functions are always positive. Thus, the sign of f'(x) depends on (-x+3). When x is less than 3, (-x+3) is positive, so f'(x) > 0.
If x > 3, both x^2 and (e^-x) are positive, and (-x+3) would be negative. Hence, f'(x) < 0.
Therefore, the function f(x) is decreasing for x < 0, and increasing for 0 < x < 3, and decreasing again for x > 3.
c) To find the inflection points of f, we need to identify where the second derivative changes sign. The second derivative is f''(x) = x(e^-x)(-6x+x^2+6).
Setting f''(x) equal to zero, we need to solve the quadratic equation:
x(e^-x)(-6x+x^2+6) = 0
Solving this equation using a graphing calculator or the quadratic formula, we find the following inflection points:
x = 0, x ≈ 4.7321, x ≈ 1.2679
d) To determine the intervals where f is concave up or concave down, we examine the sign of the second derivative f''(x).
As f''(x) = x(e^-x)(-6x+x^2+6), we see that when x < 0, f''(x) > 0 because both x and (e^-x) are negative.
Between 0 and ≈ 1.2679, we know that (e^-x) > 0. Hence, the concavity of f depends on (-6x + x^2 + 6), which changes sign at x ≈ 1.2679. Therefore, f is concave up for x < 1.2679 and concave down for 1.2679 < x < 0.
For x > ≈ 1.2679, both x and (e^-x) are positive, and the factor (-6x + x^2 + 6) is positive. So, f''(x) > 0 for x > ≈ 1.2679, indicating f is concave up.
e) To determine the asymptotic behavior of f(x), we can take the limit as x approaches positive or negative infinity. This can be done by examining the growth rate of each term.
As x approaches negative infinity, the term x^3 dominates, and the exponential term (e^-x) approaches zero. Therefore, f(x) approaches negative infinity as x approaches negative infinity.
As x approaches positive infinity, the exponential term (e^-x) dominates, and the term x^3 approaches zero. Hence, f(x) approaches zero as x approaches positive infinity.
Using limits, we can justify that f(x) has a horizontal asymptote at y = 0 as x approaches infinity.