Consider the equation below. (Give your answers correct to two decimal places.)
f(x) = 8sin(x) + 8cos(x)
0 ≤ x ≤ 2
(a) Find the intervals on which f is increasing. (Enter the interval that contains smaller numbers first.)
( , ) union ( , )
Find the interval on which f is decreasing.
( , )
(b) Find the local minimum and maximum values of f.
(min)
(max)
(c) Find the inflection points.
( , ) (smaller x value)
( , ) (larger x value)
Find the interval on which f is concave up.
( , )
Find the intervals on which f is concave down. (Enter the interval that contains smaller numbers first.)
To find the intervals on which f is increasing, we need to find the intervals where the derivative of f(x) is positive.
(a) First, let's find the derivative of f(x):
f'(x) = (8cos(x)) - (8sin(x))
= 8(cos(x) - sin(x))
Now, let's find the critical points by setting f'(x) = 0:
8(cos(x) - sin(x)) = 0
We can solve this equation by setting cos(x) = sin(x):
cos(x) = sin(x)
The solutions to this equation are x = π/4 and x = 5π/4.
Now, we can determine the intervals on which f is increasing by analyzing the sign of the derivative in each interval:
- For x < π/4, f'(x) is negative.
- For π/4 < x < 5π/4, f'(x) is positive.
- For x > 5π/4, f'(x) is negative.
Therefore, the intervals on which f is increasing are (π/4, 5π/4).
To find the interval on which f is decreasing, we can consider the intervals outside of (π/4, 5π/4). Since there are no other intervals given in the problem, we can say that f is decreasing in the interval (0, π/4) union (5π/4, 2).
(b) To find the local minimum and maximum values of f, we need to find the critical points and the endpoints of the given interval.
The critical points we found earlier are x = π/4 and x = 5π/4.
The endpoints of the given interval are x = 0 and x = 2.
Now, we need to evaluate f(x) at each of these points:
f(0) = 8sin(0) + 8cos(0) = 8(0) + 8(1) = 8
f(π/4) = 8sin(π/4) + 8cos(π/4) = 8(√2/2) + 8(√2/2) = 8√2
f(5π/4) = 8sin(5π/4) + 8cos(5π/4) = 8(-√2/2) + 8(-√2/2) = -8√2
f(2) = 8sin(2) + 8cos(2) ≈ -4.08
Therefore, the local minimum value of f is -8√2 at x = 5π/4, and the local maximum value of f is approximately 8√2 at x = π/4.
(c) To find the inflection points, we need to find where the concavity changes. This occurs when the second derivative of f(x) changes sign.
First, let's find the second derivative of f(x):
f''(x) = -8sin(x) - 8cos(x)
Setting f''(x) = 0, we can find the inflection points:
-8sin(x) - 8cos(x) = 0
sin(x) + cos(x) = 0
(sin(x) + cos(x)) / √2 = 0
sin(x + π/4) = 0
The solutions to this equation are x = -π/4, x = 3π/4, x = 5π/4, and x = 7π/4.
Therefore, the inflection points are (-π/4, 3π/4) and (5π/4, 7π/4).
To find the interval on which f is concave up, we need to find where the second derivative is positive. Evaluating f''(x) in the given interval, we find that f''(x) > 0 for x < -π/4 and 3π/4 < x < 5π/4.
To find the intervals on which f is concave down, we need to find where the second derivative is negative. Evaluating f''(x) in the given interval, we find that f''(x) < 0 for -π/4 < x < 3π/4 and 5π/4 < x < 7π/4.
Therefore, the interval on which f is concave up is (-∞, -π/4) union (3π/4, 5π/4), and the intervals on which f is concave down are (-π/4, 3π/4) and (5π/4, ∞).
To find the intervals on which a function is increasing or decreasing, we need to analyze the first derivative of the function. The function f(x) is given by f(x) = 8sin(x) + 8cos(x).
(a) To find the intervals on which f is increasing or decreasing, we need to find the critical points by setting the first derivative equal to zero and solving for x.
First, let's find the first derivative of f(x). The derivative of sin(x) is cos(x), and the derivative of cos(x) is -sin(x).
f'(x) = 8cos(x) - 8sin(x)
Setting f'(x) = 0:
0 = 8cos(x) - 8sin(x)
Now, solve for x:
8cos(x) = 8sin(x)
cos(x) = sin(x)
Since cos(x) = sin(x), we know that x must be either π/4 or 5π/4.
Now, we can determine the intervals on which f is increasing and decreasing. We can use a test point between each critical point and evaluate the sign of the first derivative in that interval to determine if it is increasing or decreasing.
For x < π/4, we can choose an arbitrary test point, say x = 0. Plug this value into the first derivative:
f'(0) = 8cos(0) - 8sin(0) = 8(1) - 8(0) = 8
Since the first derivative is positive in this interval, f is increasing.
For π/4 < x < 5π/4, we can choose another test point, say x = π/2. Plug this value into the first derivative:
f'(π/2) = 8cos(π/2) - 8sin(π/2) = 8(0) - 8(1) = -8
Since the first derivative is negative in this interval, f is decreasing.
For x > 5π/4, we can choose another test point, say x = 2. Plug this value into the first derivative:
f'(2) = 8cos(2) - 8sin(2) ≈ -5.97
Since the first derivative is negative in this interval, f is decreasing.
Therefore, the intervals on which f is increasing are (0, π/4) and (5π/4, 2).
The interval on which f is decreasing is (π/4, 5π/4).
(b) To find the local minimum and maximum values of f, we need to analyze the critical points and the endpoints of the given interval.
The critical points we found earlier are π/4 and 5π/4. We need to evaluate f(x) at these critical points as well as at the endpoints x = 0 and x = 2 to determine the local minimum and maximum values.
f(0) = 8sin(0) + 8cos(0) = 8(0) + 8(1) = 8
f(π/4) = 8sin(π/4) + 8cos(π/4) ≈ 7.07 + 7.07 ≈ 14.14
f(5π/4) = 8sin(5π/4) + 8cos(5π/4) ≈ -7.07 - 7.07 ≈ -14.14
f(2) = 8sin(2) + 8cos(2) ≈ -4.75 - 11.39 ≈ -16.14
From these calculations, we can see that the local minimum value is approximately -16.14 at x = 2, and the local maximum value is approximately 14.14 at x = π/4.
(c) To find the inflection points, we need to analyze the concavity of the function. We do this by examining the second derivative of f(x).
The second derivative of f(x) can be found by differentiating the first derivative f'(x) with respect to x:
f''(x) = -8sin(x) - 8cos(x)
To find the inflection points, we set the second derivative equal to zero and solve for x:
0 = -8sin(x) - 8cos(x)
sin(x) = -cos(x)
tan(x) = -1
From this equation, we get x = -π/4, 3π/4.
Now, we can determine the intervals on which f is concave up and concave down. We can use a test point between each inflection point and evaluate the sign of the second derivative in that interval to determine the concavity.
For x < -π/4, we can choose an arbitrary test point, say x = -π/2. Plug this value into the second derivative:
f''(-π/2) = -8sin(-π/2) - 8cos(-π/2) = -8(-1) - 8(0) = 8
Since the second derivative is positive in this interval, f is concave up.
For -π/4 < x < 3π/4, we can choose another test point, say x = 0. Plug this value into the second derivative:
f''(0) = -8sin(0) - 8cos(0) = 0 - 8(1) = -8
Since the second derivative is negative in this interval, f is concave down.
For x > 3π/4, we can choose another test point, say x = π. Plug this value into the second derivative:
f''(π) = -8sin(π) - 8cos(π) = 0 - 8(-1) = -8
Since the second derivative is negative in this interval, f is concave down.
Therefore, the interval on which f is concave up is ( -π/4, 3π/4).
The intervals on which f is concave down are ( -∞, -π/4) and ( 3π/4, ∞).