f(x) = x^2 / (x - 7)^2
(a) Find the vertical and horizontal asymptotes.
x=
y=
(b) Find the interval where the function is increasing.
Find the intervals where the function is decreasing. (Enter the interval that contains smaller numbers first.)
(c) Find the local minimum value.
(d) Find the inflection point.
Find the interval where the function is concave up. (Enter the interval that contains smaller numbers first.)
Find the intervals where the function is concave down.
(a) To find the vertical asymptote, we set the denominator equal to zero and solve for x:
x - 7 = 0
x = 7
Therefore, the vertical asymptote is x = 7.
To find the horizontal asymptote, we compare the degrees of the numerator and denominator. Since the degrees are the same (degree 2), we divide the leading coefficients:
(1/1) = 1
Therefore, the horizontal asymptote is y = 1.
(a) Vertical asymptote: x = 7
Horizontal asymptote: y = 1
(b) To find the intervals where the function is increasing, we need to find the intervals where the derivative of the function is positive.
First, let's find the derivative:
f'(x) = (2x(x - 7)^2 - 2x^2(2(x - 7)))/(x - 7)^4
= (2x(x - 7)^2 - 4x^2(x - 7))/(x - 7)^4
= (2x(x^2 - 14x + 49) - 4x^3 + 28x^2)/(x - 7)^4
= (2x^3 - 28x^2 + 98x - 4x^3 + 28x^2)/(x - 7)^4
= (-2x^3 + 98x)/(x - 7)^4
= -2x(x^2 - 49)/(x - 7)^4
To find when f'(x) > 0, we need to determine the sign of the factors.
1. x^2 - 49 > 0
Using the difference of squares, we can factor the numerator further:
(x - 7)(x + 7) > 0
The sign of (x - 7)(x + 7) is positive when x < -7 or x > 7.
2. -2x < 0
x > 0
Combining the conditions, we have two intervals where the function is increasing: (-∞, -7) U (7, ∞).
To find the intervals where the function is decreasing, we just need to find the complement of the intervals where the function is increasing. Therefore, the function is decreasing in the interval (-7, 7).
(c) To find the local minimum value, we need to find the critical points of the function.
Setting f'(x) = 0:
-2x(x^2 - 49)/(x - 7)^4 = 0
This gives us two critical points:
1. x = 0
2. x = 7
To find the y-values at these critical points, we substitute them into the original function:
1. f(0) = (0^2)/(0 - 7)^2 = 0
2. f(7) = (7^2)/(7 - 7)^2 = 49/0
From the original function, we can see that the function approaches infinity as x approaches 7. Therefore, there is no local minimum value.
(d) To find the inflection point, we need to find the places where the concavity changes.
The second derivative of the function is f''(x):
f''(x) = -2(x^2 - 49)/(x - 7)^4
To determine the concavity, we look at the sign of f''(x):
1. x^2 - 49 > 0
(x - 7)(x + 7) > 0
The sign of (x - 7)(x + 7) is negative when -7 < x < 7.
Therefore, the function is concave up in the interval (-∞, -7) U (7, ∞), and concave down in the interval (-7, 7).
Summary:
(a) Vertical asymptote: x = 7
Horizontal asymptote: y = 1
(b) Increasing interval: (-∞, -7) U (7, ∞)
Decreasing interval: (-7, 7)
(c) No local minimum value
(d) Inflection point: (-7, 7)
Concave up interval: (-∞, -7) U (7, ∞)
Concave down interval: (-7, 7)
To answer these questions, we need to analyze the properties of the given function f(x) = x^2 / (x - 7)^2 step by step:
(a) The vertical asymptote(s) can be found by checking where the denominator (x - 7)^2 equals zero, which is when x = 7. So the vertical asymptote is x = 7.
To find the horizontal asymptote, we compare the degrees of the polynomials in the numerator and denominator. In this case, both the numerator and denominator have a degree of 2. Hence, we divide the leading coefficients of the numerator and the denominator, which are both 1, to get the horizontal asymptote.
Therefore, the horizontal asymptote is given by y = 1.
(b) To determine the intervals where the function is increasing or decreasing, we look for the critical points. Critical points occur when the derivative of the function is zero or undefined.
First, let's find the derivative of f(x):
f'(x) = (2x(x - 7)^2 - 2x^2(2(x - 7)) / (x - 7)^4
Simplifying, we get:
f'(x) = (x^2 - 14x + 49) / (x - 7)^3
To find the increasing and decreasing intervals, we need to solve the inequality f'(x) > 0 for increasing and f'(x) < 0 for decreasing.
For f'(x) > 0:
x^2 - 14x + 49 > 0
(x - 7)(x - 7) > 0
(x - 7)^2 > 0
As the square of a real number is always positive, (x - 7)^2 > 0 is always true, which means the function is increasing for all x.
Hence, the function is increasing for all values of x.
(c) Since the function is increasing for all values of x, it does not have a local minimum.
(d) To find the inflection point, we need to analyze the concavity of the function. The concavity changes when the second derivative of the function changes sign or is undefined.
Let's find the second derivative of f(x):
f''(x) = (2(x - 7)^3 - 3(x - 7)^2(2x - 14)) / (x - 7)^6
Simplifying, we get:
f''(x) = (2x^2 - 14x + 23) / (x - 7)^4
To find the interval where the function is concave up, we solve the inequality f''(x) > 0.
2x^2 - 14x + 23 > 0
Using the quadratic formula, we find that this quadratic equation does not have real roots. Therefore, the function does not have an interval where it is concave up.
To find the intervals where the function is concave down, we solve the inequality f''(x) < 0.
2x^2 - 14x + 23 < 0
Using the quadratic formula, we find that this quadratic equation has real roots. Let's denote these roots as x1 and x2.
So the function is concave down on the interval (x1, x2).
Hence, the interval where the function is concave down is (x1, x2).