For y= (3x)/(2e^x+e^-1), use graphing techniques to find the approximate intervals on which the function is:
a) increasing
b) decreasing
c) concave up,
d) concave down
then find any
e) local extrema
f) inflection points
for graphing help, try
http://rechneronline.de/function-graphs/
To determine the intervals where the given function is increasing, decreasing, concave up, concave down, and to identify any local extrema and inflection points, we'll need to analyze the graph of the function.
First, let's find the critical points of the function by finding where the derivative is equal to zero or undefined. The critical points will help us identify the intervals of increasing and decreasing on the graph.
Step 1: Finding the derivative of y with respect to x:
To find the derivative of y, we can use the quotient rule:
y = (3x) / (2e^x + e^(-1))
Differentiating both the numerator and denominator separately, we get:
y' = [(3)(2e^x + e^(-1)) - (3x)(2e^x)] / (2e^x + e^(-1))^2
Simplifying the derivative gives:
y' = (6e^x + 3e^(-1) - 6xe^x) / (2e^x + e^(-1))^2
Step 2: Finding the critical points:
To find the critical points, we need to solve y' = 0.
Set y' equal to zero and solve for x:
0 = (6e^x + 3e^(-1) - 6xe^x) / (2e^x + e^(-1))^2
Multiply both sides by (2e^x + e^(-1))^2 to get rid of the denominator:
0 = 6e^x + 3e^(-1) - 6xe^x
Rearrange the equation:
6xe^x - 6e^x = -3e^(-1)
Factor out e^x:
6e^x(x - 1) = -3e^(-1)
Divide both sides by -3e^(-1):
-2(x - 1) = 1/e
Simplify:
2(x - 1) = -1/e
Distribute:
2x - 2 = -1/e
Move the constant term to the other side:
2x = -1/e + 2
Combine the terms:
2x = (2 - 1/e)
Divide both sides by 2:
x = (1 - 1/(2e))
The critical point is x ≈ 0.817.
Step 3: Analyzing the intervals:
Now let's analyze the function on the intervals to the left and right of the critical point.
1. Interval to the left of x ≈ 0.817:
Choose a value smaller than 0.817, for example, x = 0.5. Substitute it into the original equation and determine the sign of the derivative:
y' = (6e^x + 3e^(-1) - 6xe^x) / (2e^x + e^(-1))^2
y'(0.5) ≈ 0.464
Since y' > 0, the function is increasing on this interval.
2. Interval to the right of x ≈ 0.817:
Choose a value larger than 0.817, for example, x = 1.5. Substitute it into the original equation and determine the sign of the derivative:
y' = (6e^x + 3e^(-1) - 6xe^x) / (2e^x + e^(-1))^2
y'(1.5) ≈ -0.161
Since y' < 0, the function is decreasing on this interval.
Step 4: Analyzing the concavity:
To determine the concavity of the function, we need to find the second derivative:
y'' = [(6e^x + 3e^(-1) - 6xe^x)(2e^x + e^(-1))^2 - (6e^x + 3e^(-1) - 6xe^x)(2e^x + e^(-1))(2e^x - e^(-1))] / (2e^x + e^(-1))^4
Simplifying, we get:
y'' = [6e^x(2e^x + e^(-1))^2 - 6xe^x(2e^x + e^(-1))^2 + 3e^(-1)(2e^x + e^(-1))^2 - 12e^x(2e^x - e^(-1))^2] / (2e^x + e^(-1))^4
Step 5: Analyzing concavity intervals:
To identify the intervals of concavity, we need to determine the sign of y'' on either side of the critical point.
1. Interval to the left of x ≈ 0.817:
Choose a value smaller than 0.817, for example, x = 0.5. Substitute it into the second derivative, y'', and determine the sign:
y''(0.5) ≈ 2.057
Since y'' > 0, the function is concave up on this interval.
2. Interval to the right of x ≈ 0.817:
Choose a value larger than 0.817, for example, x = 1.5. Substitute it into the second derivative, y'', and determine the sign:
y''(1.5) ≈ -2.272
Since y'' < 0, the function is concave down on this interval.
Step 6: Finding local extrema:
To find the local extrema, we need to check both the critical point and the endpoints of the intervals.
At the critical point x ≈ 0.817, calculate the corresponding y-value by substituting it into the original equation:
y ≈ (3(0.817)) / (2e^(0.817) + e^(-1)) ≈ 1.461
The local maximum value is approximately y ≈ 1.461 at x ≈ 0.817.
Since we only have one critical point and the function only exists on the interval (-∞, ∞), we don't need to check the endpoints.
Step 7: Finding inflection points:
To find the inflection points, we need to check where the concavity changes.
Since we only have one critical point and it corresponds to a local maximum, there are no inflection points.
Summary:
a) The function is increasing on the interval (-∞, 0.817).
b) The function is decreasing on the interval (0.817, ∞).
c) The function is concave up on the interval (-∞, 0.817).
d) The function is concave down on the interval (0.817, ∞).
e) The local extrema are x ≈ 0.817 with a maximum value of y ≈ 1.461.
f) There are no inflection points.