Find the local maximum and minimum values of f using both the first and second derivative tests f(x) = x² / (x - 1).

Solution:

Given, f(x) = x² /(x - 1).

We can find the local maximum and minimum values using the second derivatives test.

Second derivatives test, states that when a function’s slope is zero at x, and the second derivative at x is:

1. Less than 0, it is a local maximum
2. Greater than 0, it is a local minimum
3. Equal to 0, then the test fails

Taking first derivative,

f’(x) = d(x² / (x - 1))

f’(x) = [(x-1)(2x) - (x²)(1)] / (x -1)²

f’(x) = (2x² - 2x - x²) / (x - 1)²

f’(x) = (x² - 2x) / (x - 1)²

f’(x) = [x(x - 2)] / (x - 1)²

Let f’(x) = 0

[x(x - 2)] / (x - 2)² = 0

x(x - 2) = 0

x = 0

x - 2 = 0

x = 2

x = 0 or 2

Taking second derivative,

f’’(x) = [(x² - 2x + 1)(2x - 2) - (x² - 2x) (2x - 2)] / ((x - 1)²)²

f’’(x) = [(x² - 2x + 1)(2x - 2) - (2x³ - 2x² - 4x² + 4x)] / ((x - 1)²)²

f’’(x) = [(2x³ - 2x²- 4x² + 4x + 2x - 2) - (2x³ - 2x² - 4x² + 4x)] / ((x - 1)²)²

f’’(x) = (2x - 2) / ((x - 1)²)²

f’’(x) = (2x - 2) / (x - 1)⁴

Now, f’’(x) at x = 0

f’’(0) = -2 / (-1)⁴

f’’(0) = -2 < 0

Therefore, it is a local maximum.

Now, f’’(x) at x = 2

f’’(2) = (4 - 2) / ((2 - 1)²)²

f’’(2) = 2 / 1

f''(2) = 2 > 0

Therefore, it is a local minimum.

Therefore, the local maximum and minimum values are x = 0 and x = 2.

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Find the local maximum and minimum values of f using both the first and second derivative tests f(x) = x² / (x - 1).

Summary:

The local maximum and minimum values of f(x) = x² / (x - 1) using both the first and second derivative tests is at x = 0 and x = 2.