Consider the equation below. (Round the answers to three decimal places. If you need to use -infinity or infinity, enter -INFINITY or INFINITY.)
f(x) = 9 sqrt(xe*^(-x)
(a) Find the interval on which f is increasing.
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Find the interval on which f is decreasing.
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(b) Find the local maximum value of f.
(c) Find the inflection point.
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Find the interval on which f is concave up.
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Find the interval on which f is concave down.(,)
a) 1) take the derivative set it equal to zero and you will get critical pts
2)draw a number line and pick a point to the left and right of critical points
3) plug the selected points back into the derivative and if you get a positive number write down plus and if negative write down negative over selected points. If it's going from negative to positive then it's a local minimum and decreasing on interval
(-infinity, critical point being test)
and other way around for max..
B) take the second derivative and set it equal to zero it will give you inflection pts and do the same thing again except that if it goes from - to + it will be concave up and other way around for concave down
To find the intervals on which the function f(x) = 9 sqrt(xe^(-x)) is increasing and decreasing, we need to analyze the sign of its derivative.
(a) To find the intervals where f(x) is increasing, we need to find where f'(x) > 0. First, let's find the derivative of f(x).
f(x) = 9 sqrt(xe^(-x))
f'(x) = 9 * [ (1/2) * sqrt(xe^(-x)) * (1/sqrt(xe^(-x))) * (d/dx)(xe^(-x)) ]
f'(x) = 9 * (1/2) * (1/sqrt(xe^(-x))) * ( (1 * e^(-x)) + (xe^(-x) * (-1)) )
f'(x) = 9 * (1/2) * (1/sqrt(xe^(-x))) * (e^(-x) - xe^(-x))
Simplifying further:
f'(x) = (9/2) * (1/sqrt(xe^(-x))) * e^(-x) * (1 - x)
Now, we set f'(x) > 0 to find the intervals where f(x) is increasing:
(9/2) * (1/sqrt(xe^(-x))) * e^(-x) * (1 - x) > 0
To make this equation easier to work with, we can see that (9/2) and (e^(-x) > 0 for all values of x. Therefore, we can simplify the inequality as:
(1 - x) > 0
Solving for x, we get:
1 > x
x < 1
So, the interval on which f(x) is increasing is (-∞, 1).
To find the interval on which f(x) is decreasing, we need to find where f'(x) < 0.
For f'(x) < 0, we have:
(9/2) * (1/sqrt(xe^(-x))) * e^(-x) * (1 - x) < 0
Again, considering (9/2), (e^(-x)), and (1 - x), we can simplify the inequality as:
(1 - x) < 0
Solving for x, we get:
1 < x
x > 1
So, the interval on which f(x) is decreasing is (1, ∞).
(b) To find the local maximum value of f, we can first note that f(x) is increasing on the interval (-∞, 1) and decreasing on the interval (1, ∞). Therefore, f(x) will have a local maximum at the endpoint (1).
To find the local maximum value at x = 1, we substitute x = 1 into the function:
f(1) = 9 sqrt(1 e^(-1))
f(1) = 9 sqrt(e^(-1))
f(1) ≈ 6.565
So, the local maximum value of f is approximately 6.565 at x = 1.
(c) To find the inflection point, we need to find where the concavity changes. We can do this by analyzing the sign of the second derivative, f''(x).
To find f''(x), we differentiate f'(x):
f''(x) = -9/sqrt(xe^(-x))^3 * e^(-x) * (1 - x) + 9 * (1/sqrt(xe^(-x))) * e^(-x) * (-1)
Simplifying further:
f''(x) = 9 * (-1/sqrt(xe^(-x)) * [ (1 - x) / sqrt(xe^(-x))) + 1 ]
To find the inflection point, we set f''(x) = 0:
9 * (-1/sqrt(xe^(-x)) * [ (1 - x) / sqrt(xe^(-x))) + 1 ] = 0
Simplifying this equation would be very complex, so I recommend using numerical methods or graphing software to find the exact point. However, an approximate value for the inflection point can be found by visually inspecting a graph or by estimating.
To find the intervals on which f is concave up and concave down, we can analyze the sign of f''(x).
For f''(x) > 0, the function is concave up. In other words:
9 * (-1/sqrt(xe^(-x)) * [ (1 - x) / sqrt(xe^(-x))) + 1 ] > 0
Simplifying further, we can see that the denominator (1 - x) / sqrt(xe^(-x)) would always be positive except at x = 1, where it is undefined. So, we need to focus on the sign of -1/sqrt(xe^(-x)):
-1/sqrt(xe^(-x)) > 0
Since the denominator can never be negative, we only need to consider the sign of the numerator, which is -1.
-1 > 0
This is not true, so there are no intervals on which f is concave up.
For f''(x) < 0, the function is concave down. In other words:
9 * (-1/sqrt(xe^(-x)) * [ (1 - x) / sqrt(xe^(-x))) + 1 ] < 0
Again, simplifying the inequality, we can see that the denominator is always positive except at x = 1. So, again, we only need to consider the sign of the numerator, which is -1.
-1 < 0
This is true for all x, except x = 1.
Therefore, the interval on which f is concave down is (-∞, 1) U (1, ∞).