Consider the function f(x)=xe^2x. Find the intervals on which f(x) is concave up and the intervals where it is concave down. Show a sign graph.
Please show work in detail so I can follow for future reference. Thanks!
The curve is concave up if f '' (x) is positive and
concave down if f '' (x) is negative
f ' (x) = x(2)e^(2x) + e^(2x)
= e^(2x) (2x + 1)
f '' (x) = (2x+1)(2)e^(2x) + 2 e^(2x)
= 2 e^(2x) ((2x+1) + 1)
= 2 e^(2x) (2x + 2)
= 4 e^(2x) (x+1)
since 4e^(2x) > 0 for all values of x, we just have to look at
x + 1
x+1 > 0 for x > -1
x+1 < 0 for x < -1
so the curve is concave up for all values of x > -1
and concave down for all values of x < -1
check the graph by Wolfram
http://www.wolframalpha.com/input/?i=plot+x+e%5E%282x%29%2C+-3+%3C+x+%3C+1+%2C+-.25+%3C+y+%3C+2
notice there is a point of inflection at x= -1
Oh, concave up and concave down, huh? Sounds like f(x) is going through an emotional roller coaster! Let's find those intervals.
To determine the concavity of a function, we need to look at its second derivative. So, let's start by finding the first and second derivatives of f(x).
First derivative:
f'(x) = (x)'e^2x + x(e^2x)'
= (1) e^2x + x(2e^2x)
= e^2x + 2x(e^2x)
= e^2x + 2xe^2x
= (1 + 2x)e^2x
Now, let's find the second derivative by differentiating f'(x):
Second derivative:
f''(x) = [(1 + 2x)e^2x]'
= (1 + 2x)'e^2x + (1 + 2x)(e^2x)'
= 2e^2x + (1 + 2x)(2e^2x)
= 2e^2x + 2(1 + 2x)e^2x
= 2e^2x + 2(1 + 2x)e^2x
= (2 + 4x)e^2x
Now let's find where f''(x) is equal to zero:
(2 + 4x)e^2x = 0
Since e^2x is always positive, we can solve for when (2 + 4x) = 0:
2 + 4x = 0
4x = -2
x = -1/2
So, x = -1/2 is a possible point of inflection.
Now, let's create a sign graph by considering the intervals and the sign of f''(x) within those intervals.
Interval 1: x < -1/2
Let's pick a test point, say x = -1. Plugging this value into f''(x), we get:
f''(-1) = (2 + 4(-1))e^2(-1)
= (2 - 4)e^(-2)
= -2e^(-2)
Since e^(-2) is always positive, we only need to consider the sign of -2:
-2 is negative.
Interval 2: x > -1/2
Let's again pick a test point, x = 0:
f''(0) = (2 + 4(0))e^2(0)
= 2e^0
= 2(1)
= 2
Now, let's summarize our findings:
- Interval 1: f''(x) < 0
- Interval 2: f''(x) > 0
Therefore, within the interval x < -1/2, f(x) is concave down, and within the interval x > -1/2, f(x) is concave up. But keep in mind that x = -1/2 could potentially be a point of inflection!
I hope that helps, and that f(x) gets off its emotional roller coaster soon!
To find the intervals on which the function f(x) = x•e^(2x) is concave up and concave down, we need to determine the sign of the second derivative, f''(x).
Step 1: Find the first derivative, f'(x), of the function f(x):
We'll use the product rule for differentiation and the chain rule for e^(2x).
f'(x) = (1)(e^(2x)) + (x)(2e^(2x))
= e^(2x) + 2xe^(2x)
= e^(2x)(1 + 2x)
Step 2: Find the second derivative, f''(x), of the function f(x):
We'll use the product rule for differentiation and the chain rule for e^(2x).
f''(x) = (1)(2e^(2x)) + (e^(2x))(2)
= 2e^(2x) + 2e^(2x)
= 4e^(2x)
Step 3: Determine the sign of the second derivative, f''(x), for different intervals.
The sign of f''(x) is determined by the sign of the exponential function e^(2x).
To find the intervals on which the function is concave up and concave down, we need to determine where f''(x) is positive or negative.
Setting f''(x) > 0 to find where it is positive:
4e^(2x) > 0
e^(2x) > 0
Since e^2x is always positive, there are no points where f''(x) > 0.
Setting f''(x) < 0 to find where it is negative:
4e^(2x) < 0
e^(2x) < 0
Again, since e^2x is always positive, there are no points where f''(x) < 0.
Therefore, the function f(x) = x•e^(2x) is neither concave up nor concave down for any interval.
Thus, we do not have concavity intervals or turning points to be represented on the sign graph.
To determine the intervals on which a function is concave up or concave down, we need to analyze the second derivative of the function. Let's begin by finding the first and second derivatives of the function f(x).
Given: f(x) = x * e^(2x)
Step 1: Find the first derivative, f'(x):
Using the product rule, the first derivative is obtained by differentiating the first term and keeping the second term the same, plus keeping the first term the same and differentiating the second term.
f'(x) = (x * d/dx[e^(2x)]) + (e^(2x) * d/dx[x])
Differentiating the first term (x) and keeping the second term (e^(2x)) the same:
f'(x) = 1 * e^(2x)
Differentiating the second term (e^(2x)) and keeping the first term (x) the same:
f'(x) = x * 2e^(2x)
Combining the two terms:
f'(x) = e^(2x) + 2x * e^(2x)
Step 2: Find the second derivative, f''(x):
Differentiating the first derivative obtained in Step 1 will give us the second derivative.
Using the sum rule and the product rule, we differentiate each term:
f''(x) = d/dx[e^(2x)] + d/dx[2x * e^(2x)]
Differentiating the first term (e^(2x)):
f''(x) = 2e^(2x)
Differentiating the second term (2x * e^(2x)):
f''(x) = 2 * e^(2x) + 2x * d/dx[e^(2x)]
Differentiating the last term (e^(2x)) and keeping 2x the same:
f''(x) = 2 * e^(2x) + 2x * 2e^(2x)
Combining the terms:
f''(x) = 2 * e^(2x) + 4x * e^(2x)
Now that we have the second derivative, we can analyze its sign to determine the intervals of concavity:
Step 3: Analyze the sign of f''(x):
To analyze the sign of f''(x), we set f''(x) = 0 and solve for x:
2 * e^(2x) + 4x * e^(2x) = 0
Factor out e^(2x):
e^(2x) * (2 + 4x) = 0
Setting each factor equal to zero gives:
e^(2x) = 0 (No solutions)
2 + 4x = 0
4x = -2
x = -0.5
Analyzing the sign of f''(x) in the intervals (-∞, -0.5) and (-0.5, +∞):
Choose a test point x1 < -0.5 (e.g., x1 = -1) and substitute it into f''(x):
f''(x1) = 2 * e^(2x1) + 4x1 * e^(2x1)
f''(-1) = 2 * e^(-2) + 4(-1) * e^(-2)
Calculate the value of f''(-1):
f''(-1) ≈ 2 * 0.1353 + (-4) * 0.1353 ≈ -0.1626
Since f''(-1) is negative, the function is concave down on the interval (-∞, -0.5).
Choose a test point x2 > -0.5 (e.g., x2 = 0) and substitute it into f''(x):
f''(x2) = 2 * e^(2x2) + 4x2 * e^(2x2)
f''(0) = 2 * e^0 + 4(0) * e^0
Calculate the value of f''(0):
f''(0) ≈ 2 * 1 + 0 ≈ 2
Since f''(0) is positive, the function is concave up on the interval (-0.5, +∞).
Finally, construct the sign graph:
(+∞)|======================[-0.5]======================|(-∞)
| Concave Down |
| Concave Up |
I hope this detailed explanation helps you understand the process of finding concavity intervals and constructing sign graphs. Let me know if you have any further questions!