For the function: f(x)= 2+x-x^2/ (x-1)^2 ; f'(x)= x-5/(x-1)^3 ; f''(x)=2x-14/(x-1)^4
a)find vertical and horizontal asymptotes. Examine vertical asymptote on either side of discontinuity.
b)find any local extrema
c)find points of inflection
vertical asymptotes where (x-1) = 0 : x=1
horizontal asymptote at y = -x^2/x^2 = -1
extrema where f'=0 : at x=5
inflection where f"=0 : at x=7
To find the vertical asymptotes of a function, we need to look for values of x where the denominator of the function becomes zero and the numerator does not also become zero.
a) Vertical Asymptotes:
In this case, the denominator is (x-1)^2, so the function has a vertical asymptote at x=1.
To examine the behavior of the function on either side of the vertical asymptote, we can substitute values slightly greater and slightly smaller than 1 into the function and observe the result.
- On the left side of x=1, we can choose x=0.9. Plugging this into the function, we get:
f(0.9) = 2 + 0.9 - 0.9^2 / (0.9 - 1)^2 = 1.89.
- On the right side of x=1, we can choose x=1.1. Plugging this into the function, we get:
f(1.1) = 2 + 1.1 - 1.1^2 / (1.1 - 1)^2 = 2.09.
Therefore, as x approaches 1 from both sides, the function approaches 1.89 and 2.09, respectively.
For horizontal asymptotes, we need to examine the behavior of the function as x approaches positive or negative infinity. To do this, we need to consider the degree of the numerator and denominator.
- The degree of the numerator is 1, and the degree of the denominator is 2. Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y=0.
b) Local Extrema:
To find the local extrema, we need to look for the critical points, where the derivative is either 0 or undefined. In this case, the derivative is:
f'(x) = (x-5) / (x-1)^3.
To find the critical points, we set f'(x) equal to zero and solve for x:
(x-5) / (x-1)^3 = 0
The numerator can only be zero when x=5, and the denominator is never zero. Therefore, x=5 is the only critical point.
To determine if this critical point corresponds to a local maximum or minimum, we can analyze the sign of the derivative on either side of x=5.
- For x < 5, we can choose x=4. Plugging this into the derivative, we get:
f'(4) = (4-5) / (4-1)^3 = -1 / 27 < 0.
- For x > 5, we can choose x=6. Plugging this into the derivative, we get:
f'(6) = (6-5) / (6-1)^3 = 1 / 125 > 0.
Since the derivative changes sign from negative to positive as we move from left to right of x=5, we can conclude that there is a local minimum at x=5.
c) Points of Inflection:
To find points of inflection, we need to look for values of x where the second derivative changes sign. In this case, the second derivative is:
f''(x) = (2x-14) / (x-1)^4.
To find the points of inflection, we need to set f''(x) equal to zero and solve for x:
(2x-14) / (x-1)^4 = 0
The numerator can only be zero when x=7, and the denominator is never zero. Therefore, x=7 is the only point of inflection.
To determine if this point of inflection is concave up or concave down, we can analyze the sign of the second derivative on either side of x=7.
- For x < 7, we can choose x=6. Plugging this into the second derivative, we get:
f''(6) = (2*6-14) / (6-1)^4 = -2 / 625 < 0.
- For x > 7, we can choose x=8. Plugging this into the second derivative, we get:
f''(8) = (2*8-14) / (8-1)^4 = 2 / 625 > 0.
Since the second derivative changes sign from negative to positive as we move from left to right of x=7, we can conclude that there is a point of inflection at x=7.