3. Let f be the function defined by f(x)=ln(2+sinx) for pi<=x<=2pi
a. Find the absolute maximum value and the absolute minimum value of f. Show the analysis that leads to your conclusion.
b. Find the x-coordinate of each inflection point on the graph of f. Justify your answer.
a. To find the absolute maximum and minimum values of f(x), we need to analyze the critical points and the endpoints of the given interval (pi <= x <= 2pi).
First, let's find the critical points of f(x) by taking the derivative of f(x) and setting it equal to 0:
f'(x) = d/dx [ln(2+sinx)]
= (1/(2+sinx)) * (cosx)
Setting this equal to zero:
(1/(2+sinx)) * (cosx) = 0
To find where the denominator is zero (which would result in an undefined value), we set 2+sinx = 0:
sinx = -2
However, in the given interval (pi <= x <= 2pi), sinx cannot equal -2. Therefore, there are no critical points in the interval (pi <= x <= 2pi).
Next, we check the endpoints of the interval:
f(pi) = ln(2 + sin(pi)) = ln(2 + 0) = ln(2)
f(2pi) = ln(2 + sin(2pi)) = ln(2 + 0) = ln(2)
Since ln(2) is a positive value, f(pi) and f(2pi) will have the same value. Therefore, we can conclude that ln(2) is both the absolute maximum and minimum value of f in the given interval.
b. To find the inflection points of f(x), we need to analyze the concavity of f(x) at various points. We can do this by finding the second derivative of f(x):
f''(x) = d^2/dx^2 [ln(2+sinx)]
= -((cosx)^2)/(2+sinx)
For an inflection point to occur, the second derivative must equal zero or be undefined. However, in the given interval (pi <= x <= 2pi), the second derivative is never undefined.
To find where the second derivative is zero, we set the numerator equal to zero:
(cosx)^2 = 0
This equation has no solutions within the interval (pi <= x <= 2pi). Therefore, there are no inflection points on the graph of f(x) within the given interval.
To find the absolute maximum and minimum values of the function f(x) = ln(2 + sin(x)) for π ≤ x ≤ 2π, we need to first find the critical points and endpoints of the interval.
a) Finding the critical points:
To find the critical points, we need to find where the derivative of f(x) equals zero or does not exist.
f'(x) = d/dx(ln(2 + sin(x))) = (1/(2 + sin(x)))*(cos(x))
Setting the derivative equal to zero:
(1/(2 + sin(x)))*(cos(x)) = 0
For the derivative to equal zero, the numerator cos(x) must equal zero. This occurs at x = π/2 and x = (3π)/2.
Now, we need to check these critical points, as well as the endpoints of the interval, to determine if they are a maximum or minimum. We also need to check the function f(x) where the derivative does not exist, if any.
Checking endpoints:
When x = π, f(π) = ln(2 + sin(π)) = ln(2 + 0) = ln(2) ≈ 0.693.
When x = 2π, f(2π) = ln(2 + sin(2π)) = ln(2 + 0) = ln(2) ≈ 0.693.
Checking critical points:
When x = π/2, f(π/2) = ln(2 + sin(π/2)) = ln(2 + 1) = ln(3) ≈ 1.099.
When x = (3π)/2, f((3π)/2) = ln(2 + sin((3π)/2)) = ln(2 - 1) = ln(1) = 0.
So, the absolute maximum value of f(x) occurs at x = π/2 with a value of ln(3) ≈ 1.099, and the absolute minimum value occurs at x = (3π)/2 with a value of 0.
b) Finding the inflection points:
To find the inflection points, we need to find where the second derivative changes sign or does not exist.
f''(x) = d^2/dx^2(ln(2 + sin(x))) = (-1/(2 + sin(x))^2)*(cos^2(x)) - (1/(2 + sin(x)))*(sin(x))
Setting the second derivative equal to zero:
(-1/(2 + sin(x))^2)*(cos^2(x)) - (1/(2 + sin(x)))*(sin(x)) = 0
Rearranging:
(-cos^2(x))/(2 + sin(x))^2 = sin(x)/(2 + sin(x))
Expanding and simplifying:
-sin^2(x) = cos(x)*(2 + sin(x))
Dividing both sides by cos(x):
-sin(x) = 2 + sin(x)
Bringing terms to one side:
2sin(x) = -2
Dividing by 2:
sin(x) = -1
The sin(x) = -1 occurs when x = 3π/2.
So, the x-coordinate of the inflection point on the graph of f is 3π/2.
Justification: At the inflection point, the concavity of the function changes. In this case, the second derivative changes sign from negative to positive, indicating a change in concavity from concave downward to concave upward.
To find the absolute maximum and minimum values of the function f(x) = ln(2 + sinx) for π ≤ x ≤ 2π, we can follow these steps:
a. Finding the critical points:
1. Calculate the derivative of f(x):
f'(x) = (1/(2 + sinx)) * cosx
2. Set f'(x) = 0 and solve for x:
(1/(2 + sinx)) * cosx = 0
There are no solutions for cosx = 0 in the given interval [π, 2π]. Therefore, f'(x) has no critical points in this interval.
b. Determining the endpoints:
Evaluate f(x) at the endpoints of the interval [π, 2π]:
f(π) = ln(2 + sinπ) = ln(2 + 0) = ln(2) ≈ 0.693
f(2π) = ln(2 + sin2π) = ln(2 + 0) = ln(2) ≈ 0.693
Since ln(2) is the same value at both endpoints, it is both the absolute maximum and absolute minimum value of f(x) over the interval [π, 2π].
To find the x-coordinate of each inflection point on the graph of f, we can follow these steps:
1. Calculate the second derivative of f(x):
f''(x) = -[(1/(2+sinx))^2] * sinx + (1/(2+sinx)) * cosx * cosx
2. Set f''(x) = 0 and solve for x:
-[(1/(2+sinx))^2] * sinx + (1/(2+sinx)) * cosx * cosx = 0
This equation is quite complex and nonlinear, making it challenging to find exact solutions mathematically. However, we can use numerical methods or graphing software to determine the approximate x-coordinates of the inflection points.
By analyzing the sign changes of the second derivative, we can conclude that there are at least one or more inflection points in the interval [π, 2π]. The exact x-coordinates can be further approximated using numerical methods if required.
f'(x) = -cosx/(2+sinx)
for max/min,
-cosx/(2+sinx)= 0
cosx = 0
x = π/2 or x = 3π/2
when x =π/2
f(π/2) = ln(2 + 1) = ln 3
when x = 3π/2
f(3π/2) = ln(2-1) = ln 1 = 0
so which is max and which is min ?
for pts of inflection, find
f''(x), then set that equal to zero
I suggest the quotient rule.