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Chapter 3: Problem 28

Find all relative extrema. Use the Second Derivative Test where applicable. \(f(x)=\frac{x}{x-1}\)

Short Answer

Expert verified
The function \(f(x)=\frac{x}{x-1}\) has a relative minimum at \(x = 1\).

Step by step solution

01

Differentiation

Start the process by differentiating this function with respect to \(x\). Remember to use the quotient rule. The first derivative of the function is \(f'(x) = \frac{(x-1)\cdot1 - x\cdot1}{(x-1)^2} = -\frac{1}{(x-1)^2}\)
02

Find the Critical Points

Set the first derivative equal to zero - however, in this case, that does not provide any solutions, as a fraction can only be zero if the numerator is zero, but our numerator is -1 (a constant). Here, the first derivative is undefined for \(x = 1\), which makes \(x = 1\) a critical point.
03

Second Derivative

Now, take the second derivative of \(f(x)\). The second derivative of the function is \(f''(x) = \frac{2}{(x-1)^3}\).
04

Second Derivative Test

Then, perform the second derivative test by inserting the critical point into the second derivative function. We get \(f''(1)\), which is undefined. The second derivative test fails because it can't determine the concavity of the function at \(x = 1\).
05

Analyze the Sign of the First Derivative

Due the failure of the Second Derivative Test, we now need to check sign changes in the first derivative. This can be done by creating a number line with the critical points and check our \(f'(x)\) on each interval. Choose the values around 1, like 0.5 and 1.5. For \(f'(0.5)\), we get a negative value and for \(f'(1.5)\), we get a positive value. Hence the function decreases before \(x = 1\) and increases after, which means there's a relative minimum at \(x = 1\)

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