Problem 42 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the second derivative and solve the equation \(f^{\prime \prime}(x)=0\). $$ f(x)=\frac{x}{x-1} $$

Step-by-Step Solution

Answer

There are no real number solutions for \( x \) where \( f^{\prime \prime}(x) = 0 \). However, \( x = 1 \) is a critical point as the function is undefined here.

Step 1: Find the First Derivative

To find the first derivative of the function, use the quotient rule. Recall the quotient rule states that the derivative of \( \frac{u}{v} \) is \( \frac{vu' - uv'}{v^2} \). Here, \( u = x \) and \( v = x - 1 \). So, \( u' = 1 \) and \( v' = 1 \). Applying the quotient rule gives \( f^{\prime}(x) = \frac{(x - 1) \cdot 1 - x \cdot 1}{(x - 1)^2} = \frac{-1}{(x - 1)^2} \).

Step 2: Find the Second Derivative

Finding the second derivative involves differentiating the first derivative. The first derivative is \( \frac{-1}{(x - 1)^2} \), a negative simple fraction, which can be rewritten as \( -1 \cdot (x - 1)^{-2} \). To differentiate this, use the chain rule where multiplication of the current exponent and subtraction of one from the current exponent need to be applied. The chain rule is applied due to the presence of the function \( x - 1 \) inside the power function. Hence, \( f^{\prime \prime}(x) = -1 \cdot -2 \cdot (x - 1)^{-3} = \frac{2}{(x - 1)^3} \).

Step 3: Solve \( f^{\prime \prime}(x) = 0 \)

Setting the second derivative equal to zero and solving for \( x \) gives: \( 0 = \frac{2}{(x - 1)^3} \). Retail that any number divided by zero is undefined, so there are no real solutions but a critical point at \( x = 1 \) exists because the function \( f^{\prime \prime}(x) \) is undefined there. It may be a point of inflection or possible extremum point.

Step 4: Conclusion

Although there are no specific \( x \) value solutions for which the second derivative equals zero, one must be aware that \( x = 1 \) is a critical point. The function \( f^{\prime \prime}(x) \) is undefined at \( x = 1 \).

Key Concepts

Quotient Rule

When you come across a function expressed as a fraction or ratio, like \( \frac{x}{x-1} \), to differentiate it, we use the Quotient Rule. This rule is a powerful tool that tells us how to take the derivative of a quotient. Here's the basic form: for a function \( \frac{u}{v} \), the derivative is given by \( \frac{vu' - uv'}{v^2} \). What this means is that you:

• Differentiate the top part (numerator) of the fraction, called \( u \), which becomes \( u' \).
• Do the same for the bottom part (denominator), \( v \), called \( v' \).
• Multiply the original bottom by the derivative of the top and subtract the product of the original top and the derivative of the bottom.
• Finally, divide everything by the square of the bottom part.

In our function \( \frac{x}{x-1} \), \( u = x \) and \( v = x - 1 \). Plugging these into the Quotient Rule helps find the first derivative \( f'(x) = \frac{-1}{(x - 1)^2} \). This provides the foundation needed to find the second derivative in the next steps.

Chain Rule

After applying the Quotient Rule, we get a function \( \frac{-1}{(x - 1)^2} \) that can also be expressed as \( -1 \cdot (x-1)^{-2} \). To find the second derivative, you'll use the Chain Rule. This rule is handy when dealing with composite functions, which are functions composed of other functions.The Chain Rule tells us how to differentiate a composite function, like a function raised to a power. It involves:

• Taking the derivative of the outer function (in our case, related to the power \(-2\)).
• Multiplying this by the derivative of the inner function (here, \( x - 1 \)).

For example, the derivative of \( (x - 1)^{-2} \) is \(-2(x - 1)^{-3}\). By applying the Chain Rule, we get \( -1 \cdot -2 \cdot (x-1)^{-3}\), simplifying to \( \frac{2}{(x-1)^3} \) as the second derivative \( f''(x) \). Understanding and applying the Chain Rule allows us to navigate through complex differentiation problems with ease.

Critical Points

After finding the second derivative, \( f''(x) = \frac{2}{(x-1)^3} \), a student might be tempted to set it equal to zero to find critical points because that's usually a method to find potential points of curve changes.However, in this case, setting \( \frac{2}{(x-1)^3} = 0 \) leads to no real solution because a non-zero number cannot be zero.
But this doesn’t mean the function doesn’t have a critical point – it’s just different here. The difficulty and uniqueness of this exercise lies in understanding that if the second derivative is undefined at \( x = 1 \), then \( x = 1 \) is still a critical point.This makes \( x = 1 \) a location where the behavior of the function could change, like an inflection point or vertical tangent. Being aware of different types of critical points is key in calculus, helping you comprehend where functions may have extrema or change concavity.