Suppose that f(x)=x−7x^[1/7]
(A) Find all critical values of f.
(B) Use interval notation to indicate where f(x) is increasing and also decreasing.
(C) Find the x-coordinates of all local maxima and local minima of f.
(D) Use interval notation to indicate where f(x) is concave up and also concave down.
(E) Find all inflection points of f.
To answer these questions step-by-step, let's start with finding the derivative of f(x) in order to determine the critical values.
Step 1: Find the derivative of f(x):
f'(x) = 1 - 7/7 * x^(-6/7) = 1 - x^(-6/7)
Step 2: Set the derivative equal to zero to find critical values:
1 - x^(-6/7) = 0
x^(-6/7) = 1
Taking the 7/6 root of both sides:
x = 1^(7/6)
x = 1
(A) The critical value of f is x = 1.
Step 3: Determine intervals of increase and decrease using the first derivative test.
To do this, we need to examine the sign of the derivative in different intervals:
For x < 1:
Choose a value in this interval, for example, x = 0.
f'(0) = 1 - 0 < 0
Therefore, f(x) is decreasing for x < 1.
For x > 1:
Choose a value in this interval, for example, x = 2.
f'(2) = 1 - 2^(-6/7) > 0
Therefore, f(x) is increasing for x > 1.
(B) In interval notation, f(x) is decreasing for x < 1 and increasing for x > 1: (-∞, 1) and (1, ∞).
Step 4: Find the second derivative of f(x) to determine concavity.
Step 4.1: Differentiate f'(x):
f''(x) = (1 - x^(-6/7))' = 0 + 6/7 * x^(-6/7 - 1) = 6/7 * x^(-13/7)
Step 4.2: Set f''(x) equal to zero to find points of inflection:
6/7 * x^(-13/7) = 0
x^(-13/7) = 0
Since x^(-13/7) cannot be equal to zero, there are no points of inflection.
(C) There are no local maxima or minima of f(x).
Step 5: Determine intervals of concavity using the second derivative test.
For x < 1:
Choose a value in this interval, for example, x = 0.
f''(0) = 6/7 * 0^(-13/7) = undefined
Since f''(x) is undefined, we cannot determine the concavity in this interval.
For x > 1:
Choose a value in this interval, for example, x = 2.
f''(2) = 6/7 * 2^(-13/7) > 0
Therefore, f(x) is concave up for x > 1.
(D) In interval notation, f(x) is concave up for x > 1: (1, ∞).
(E) There are no inflection points for f(x).
To solve each part of the question, we need to find the derivative of the function and analyze its behavior.
(A) To find all critical values of f, we need to find where the derivative is equal to zero or undefined.
First, let's find the derivative of f(x):
f'(x) = 1 - 7(1/7)x^(1/7 - 1)
Simplifying the derivative:
f'(x) = 1 - x^(-6/7)
Setting the derivative equal to zero and solving for x:
1 - x^(-6/7) = 0
x^(-6/7) = 1
x^(6/7) = 1
Taking the 7th power of both sides:
x^6 = 1
x = ±1
So the critical values of f are x = 1 and x = -1.
(B) To determine where f(x) is increasing or decreasing, we need to analyze the sign of the derivative.
When f'(x) > 0, f(x) is increasing.
When f'(x) < 0, f(x) is decreasing.
Let's determine the intervals where f(x) is increasing and decreasing.
Using test points between the critical values and beyond, we can construct a sign chart for f'(x):
Test point: x = 0
f'(0) = 1 - 0^(-6/7) = 1 - 0 = 1
Sign Chart for f'(x):
-----------------------
x | f'(x)
-----------------------
-∞ | +
-1 | -
0 | +
1 | -
From the sign chart, we can see that f(x) is increasing on (-∞, -1) and (0, ∞), and decreasing on (-1, 0).
(C) To find the x-coordinates of all local maxima and local minima of f, we need to analyze the critical points.
To classify these critical points as local maxima or minima, we can examine the behavior of the function around the critical points.
Using the second derivative test:
f''(x) = (6/7)x^(-6/7 - 1)
f''(x) = (6/7)x^(-13/7)
For x = 1:
f''(1) = (6/7)(1)^(-13/7) = 6/7
For x = -1:
f''(-1) = (6/7)(-1)^(-13/7) = -6/7
Since f''(1) > 0, we have a local minimum at x = 1.
Since f''(-1) < 0, we have a local maximum at x = -1.
(D) To determine where f(x) is concave up or concave down, we need to analyze the sign of the second derivative.
When f''(x) > 0, f(x) is concave up.
When f''(x) < 0, f(x) is concave down.
Using the information from part (C), we can construct a sign chart for f''(x):
Sign Chart for f''(x):
-----------------------
x | f''(x)
-----------------------
-∞ | -
-1 | -
1 | +
From the sign chart, we can see that f(x) is concave down on (-∞, -1) and concave up on (-1, ∞).
(E) To find the inflection points of f, we need to find where the concavity changes.
From part (D), we know that the concavity changes at x = -1.
Therefore, the inflection point of f is at x = -1.
f' = 1 - 1/x^(6/7)
so, critical values are where x=0 or x^(6/7) = 1
f is increasing where f' > 0
max/min are at critical values
concave up where f" > 0
inflection points where f" = 0