f(x)= x^2 / (x-2)^2
find the intervals of concavity and the inflections points
I will be happy to critique your work.
If the function is twice differentiable within its domain, inflection points occur at values of c where f"(c)=0.
Note that not all points where f"(c)=0 are inflection points.
You need to check the concavity between points where f"(x)=0. It is concave up if f"(x)>0 and concave down if f"(x)<0.
X=c is an inflection point if all of the following conditions are satisfied:
1. f(x) is twice differentiable at c.
2. f"(c)=0
3. f"(c+) has a different sign than f"(c-), i.e. concavity changes.
Now it's time for you to sharpen your pencil and show us some work.
To find the intervals of concavity and the inflection points of the function f(x) = x^2 / (x-2)^2, we need to analyze the second derivative of the function. Let's begin by finding the first derivative, then the second derivative.
Step 1: Find the first derivative of f(x):
To find the first derivative, we can use the quotient rule. Recall that if we have a function u(x) / v(x), the derivative can be found using the formula:
(f/g)' = (f'g - fg') / g^2
Applying the quotient rule to our function f(x) = x^2 / (x-2)^2, we get:
f'(x) = [(2x(x-2)^2) - (x^2 * 2(x - 2))] / (x - 2)^4
Simplifying further:
f'(x) = (2x(x-2)^2 - 2x^2(x-2)) / (x-2)^4
= (2x(x-2)(x-2 - x)) / (x-2)^4
= (2x(x-2)(x-2 - x)) / (x-2)^4
= - (2x(x-2)) / (x-2)^3
= -2(x(x-2)) / (x-2)^3
= -2x / (x-2)
Step 2: Find the second derivative of f(x):
Now, let's differentiate the first derivative we obtained in Step 1 to find the second derivative.
f''(x) = d/dx (-2x / (x-2))
Using the quotient rule, we have:
f''(x) = [(2(x-2)) - (1(-2x))] / (x - 2)^2
= (2x - 4 + 2x) / (x - 2)^2
= (4x - 4) / (x - 2)^2
Step 3: Determine the intervals of concavity:
To find the intervals of concavity, we need to examine the sign of the second derivative. Since the denominator, (x - 2)^2, is always positive, the sign of the second derivative depends on the numerator.
Setting the numerator equal to zero:
4x - 4 = 0
4x = 4
x = 1
We have found a critical value at x = 1. This critical value divides our number line into two intervals: (-∞, 1) and (1, ∞).
For x < 1, the numerator 4x - 4 is negative, meaning the second derivative f''(x) is negative. So, the function is concave down in the interval (-∞, 1).
For x > 1, the numerator 4x - 4 is positive, meaning the second derivative f''(x) is positive. So, the function is concave up in the interval (1, ∞).
Step 4: Find the inflection point(s):
An inflection point occurs where the concavity changes. So, we can find the inflection point by determining when the second derivative f''(x) changes sign.
Substituting a value on either side of the critical value into the second derivative:
For x = 0 (the interval (-∞, 1)):
f''(0) = (4(0) - 4) / (0 - 2)^2
= -4 / 4
= -1 (negative)
For x = 2 (the interval (1, ∞)):
f''(2) = (4(2) - 4) / (2 - 2)^2
= 4 / 0
= undefined
Since the second derivative is undefined at x = 2, it means that there is no inflection point at x = 2.
From the analysis, we have an inflection point at x = 0 since the second derivative f''(0) changes sign from negative to positive.
To summarize:
- The function f(x) = x^2 / (x-2)^2 is concave down in the interval (-∞, 1).
- The function f(x) = x^2 / (x-2)^2 is concave up in the interval (1, ∞).
- The function has an inflection point at x = 0.