Problem 42 - Calculus an Applied Approach

Question

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 solutions for the equation \(f''(x) = 0\).

Step 1: Differentiate the function to find f'(x)

To find the first derivative, use the quotient rule which states \((\frac{u}{v})' = \frac{vu' - uv'}{v^2}\) where in this scenario, \(u = x\) and \(v = x - 1\). So, \(f'(x) = \frac{(x-1) - x}{(x-1)^2} = \frac{-1}{(x-1)^2}\)

Step 2: Find the Second Derivative f''(x)

Apply the chain rule to find the second derivative as the function \(f'(x)\) is itself a quotient. \((f'(x))' = f''(x) = ([-\frac{1}{(x-1)^2}])' = \frac{2}{(x-1)^3}\)

Step 3: Solve the Equation f''(x) = 0

To find the critical points, solve the equation \(f''(x) = 0\). Setting \(\frac{2}{(x-1)^3} = 0\) and solving results in an equation with no solutions. The denominator cannot be 0, therefore \(x \neq 1\). The numerator determines the zero value, however, it's a constant and thus, there are no solutions for x in the given equation.

Key Concepts

Second Derivative

In calculus, the second derivative provides us with vital information about the concavity and the inflection points of a function. To find the second derivative, first, find the first derivative of the function. For the function given, \(f(x)=\frac{x}{x-1}\), we initially differentiate using the quotient rule, resulting in \(f'(x) = \frac{-1}{(x-1)^2}\). The second derivative \(f''(x)\) follows by differentiating \(f'(x)\). This second derivative can reveal critical points by setting it to zero and solving the resulting equation. However, as we found in the solution, there were no real solutions for \(f''(x)=0\) in this particular case. This means the function does not have points of inflection. Understanding the second derivative is crucial because it shows how the rate of change (determined by the first derivative) itself changes over an interval.

Quotient Rule

The quotient rule is a fundamental tool in calculus for differentiating functions that are expressed as a quotient of two other functions. The rule can be stated as: given a function \(\frac{u}{v}\), its derivative is \(\left(\frac{u}{v}\right)' = \frac{v u' - u v'}{v^2}\). For the function \(f(x)=\frac{x}{x-1}\), we identified \(u=x\) and \(v=x-1\). Calculating their derivatives, we have \(u'=1\) and \(v'=1\). Applying the quotient rule leads to:

• Substitute: \(f'(x) = \frac{(x-1)(1) - x(1)}{(x-1)^2}\)
• Simplify: \(f'(x) = \frac{-1}{(x-1)^2}\)

The quotient rule is invaluable when dealing with rational functions because it simplifies the process of differentiation significantly. It's important to carefully apply the rule step by step, as it's easy to make mistakes with signs or order of operations.

Chain Rule

The chain rule is another key concept in calculus used to differentiate composite functions. It states that if you have a composition of two functions, say \(g(h(x))\), the derivative is \((g(h(x)))' = g'(h(x))\cdot h'(x)\). In our example, after finding \(f'(x)\), the function itself was a quotient requiring further differentiation. When moving from \(f'(x) = \frac{-1}{(x-1)^2}\) to find \(f''(x)\), the systemic application of the chain rule allows us to address the nested layer of differentiation. We considered the inner function \((x-1)^2\) and its derivative \(h'(x)\), while \(g(x) = -1/h(x)\) necessitated the outside differentiation.

• Inner function: \((x-1)^2\) leading to \(h'(x) = 2(x-1)\)
• Outer function: \(-1/u\) leading to \(g'(u) = \frac{2}{(x-1)^3}\)

The chain rule is invaluable when expressions are layered, ensuring you can unravel even the most complex derivatives seamlessly.