2. Consider the function f defined by f(x)=(e^X)cosx with domain[0,2pie] .
a. Find the absolute maximum and minimum values of f(x)
b. Find the intervals on which f is increasing.
c. Find the x-coordinate of each point of inflection of the graph of f.
first, it's pi, not pie
f = e^x cosx
f' = e^x (cosx - sinx)
f'' = -2e^x sinx
max/min where f' = 0 and f'' not zero
f increasing where f' > 0
inflection at f'' = 0
Come back if you get stuck and show us what happened.
a. To find the absolute maximum and minimum values of f(x), we need to evaluate the function at the critical points and endpoints.
First, let's find the critical points by taking the derivative of f(x) and setting it equal to zero:
f'(x) = e^x * cos(x) - e^x * sin(x) = 0
Factoring out e^x, we get:
e^x * (cos(x) - sin(x)) = 0
This equation is satisfied when either e^x = 0 or cos(x) - sin(x) = 0. However, e^x is always positive, so we can disregard the first factor.
Now, let's solve the equation cos(x) - sin(x) = 0:
cos(x) = sin(x)
Divide both sides by cos(x):
1 = tan(x)
x = arctan(1)
x = π/4 (since 0 ≤ x ≤ 2π)
Now, we need to evaluate f(x) at the critical point and endpoints:
f(0) = e^0 * cos(0) = 1
f(2π) = e^(2π) * cos(2π) = e^(2π) * 1 = e^(2π)
Therefore, the absolute minimum value is 1 at x = 0, and the absolute maximum value is e^(2π) at x = 2π.
b. To find the intervals on which f is increasing, we need to find where the derivative is positive.
From part a, we found that f'(x) = e^x * (cos(x) - sin(x)). Since e^x is always positive, we only need to consider the sign of (cos(x) - sin(x)).
To determine the intervals where f is increasing, we need to find where (cos(x) - sin(x)) > 0.
We can analyze the sign of cos(x) - sin(x) by considering the values of cos(x) and sin(x) in each quadrant of the unit circle.
In the first quadrant (0 ≤ x ≤ π/2), cos(x) and sin(x) are both positive, so cos(x) - sin(x) is positive.
In the second quadrant (π/2 ≤ x ≤ π), cos(x) is negative and sin(x) is positive, so cos(x) - sin(x) is negative.
In the third quadrant (π ≤ x ≤ 3π/2), cos(x) and sin(x) are both negative, so cos(x) - sin(x) is positive.
In the fourth quadrant (3π/2 ≤ x ≤ 2π), cos(x) is positive and sin(x) is negative, so cos(x) - sin(x) is positive.
Therefore, f is increasing on the intervals [0, π/2) and (π, 3π/2].
c. To find the x-coordinate of each point of inflection, we need to find where the second derivative of f(x) changes sign.
Taking the second derivative of f(x), we get:
f''(x) = e^x * (cos(x) - sin(x)) - e^x * (sin(x) + cos(x))
Simplifying, we get:
f''(x) = e^x * (2cos(x) - 2sin(x))
To find where f''(x) changes sign, we need to set f''(x) equal to zero:
e^x * (2cos(x) - 2sin(x)) = 0
This equation is satisfied when either e^x = 0 or 2cos(x) - 2sin(x) = 0. However, e^x is always positive, so we can disregard the first factor.
To solve 2cos(x) - 2sin(x) = 0:
cos(x) = sin(x)
Divide both sides by cos(x):
1 = tan(x)
x = arctan(1)
x = π/4
Therefore, the x-coordinate of each point of inflection is π/4.
a. To find the absolute maximum and minimum values of the function f(x) = e^x * cos(x) on the domain [0, 2π], we can use the properties of continuous functions and critical points.
1. Start by finding the critical points of f(x) by taking the derivative f'(x):
f'(x) = (e^x * cos(x))' = (e^x * cos(x)) - (e^x * sin(x))
2. Set f'(x) = 0 to find the critical points:
(e^x * cos(x)) - (e^x * sin(x)) = 0
3. Simplify the equation:
cos(x) - sin(x) = 0
4. Solve for x by setting cos(x) = sin(x):
cos(x)/sin(x) - 1 = 0
cot(x) - 1 = 0
cot(x) = 1
5. The solutions for cot(x) = 1 are x = π/4 and x = 5π/4.
6. Now, evaluate f(x) at the critical points and endpoints of the domain [0, 2π]:
f(0) = (e^0) * cos(0) = 1 * 1 = 1
f(2π) = (e^(2π)) * cos(2π) = 1 * 1 = 1
f(π/4) ≈ (e^(π/4)) * cos(π/4)
f(5π/4) ≈ (e^(5π/4)) * cos(5π/4)
7. Compare the values of f(x) at the critical points and endpoints to find the absolute maximum and minimum values.
b. To find the intervals on which f(x) is increasing or decreasing, we can examine the sign of the derivative f'(x) on various intervals.
1. Recall that f'(x) = (e^x * cos(x)) - (e^x * sin(x)).
2. Find the critical points of f(x) as found in part (a): x = π/4, x = 5π/4.
3. Determine the sign of f'(x) in the intervals (-∞, π/4), (π/4, 5π/4), and (5π/4, ∞).
- Plug in a value from each interval into f'(x) to determine if it is positive or negative.
- For example, plug in x = 0 into f'(x).
- f'(0) = (e^0 * cos(0)) - (e^0 * sin(0))
- f'(0) = 1 - 0 = 1
- Since f'(0) is positive, the function is increasing on the interval (-∞, π/4).
4. Repeat this process for the remaining intervals.
c. To find the x-coordinate of each point of inflection of the graph of f(x), we need to find the second derivative f''(x) and determine where it changes sign.
1. Start by finding the second derivative f''(x) by taking the derivative of f'(x):
f''(x) = ((e^x * cos(x)) - (e^x * sin(x)))'
f''(x) = (e^x * cos(x))' - (e^x * sin(x))'
2. Simplify the equation:
f''(x) = (-e^x * sin(x)) - (e^x * cos(x))
3. Set f''(x) = 0 to find the potential points of inflection:
(-e^x * sin(x)) - (e^x * cos(x)) = 0
4. Simplify the equation:
sin(x) + cos(x) = 0
5. Solve for x by setting sin(x) = -cos(x):
sin(x)/cos(x) + 1 = 0
tan(x) + 1 = 0
tan(x) = -1
6. The solution for tan(x) = -1 is x = 3π/4.
7. Now, evaluate f''(x) at the potential point of inflection and other critical points:
f''(3π/4) = (-e^(3π/4) * sin(3π/4)) - (e^(3π/4) * cos(3π/4))
8. Check if the second derivative changes sign around the potential point of inflection and other critical points.