Problem 42 - Thomas Calculus in SI Units

Question

Let \(f(x)=x^{2}-4 x-5, x>2 .\) Find the value of \(d f^{-1} / d x\) at the point \(x=0=f(5)\).

Step-by-Step Solution

Answer

The derivative of \( f^{-1}(x) \) at \( x = 0 \) is \( \frac{1}{6} \).

Step 1: Understand the Problem

We need to find the derivative of the inverse function of \(f(x)\), denoted as \(f^{-1}(x)\), and evaluate it at a point where \(x = 0\), which is \(f(5)\). This means we want \(d f^{-1}/dx \) at \(x=0\).

Step 2: Verify the Conditions

Given, \(f(x) = x^2 - 4x - 5\) with \(x > 2\). First, confirm that \(f(5) = 0\). Calculate \(f(5) = 5^2 - 4\cdot5 - 5 = 25 - 20 - 5 = 0\). Thus, \(f(5)=0\) is true.

Step 3: Differentiate \(f(x)\)

Find the derivative of \(f(x)\). \(f'(x)\) is obtained by differentiating \(f(x) = x^2 - 4x - 5\):\[ f'(x) = 2x - 4 \].

Step 4: Apply the Inverse Function Derivative Formula

The derivative of an inverse function, \(g = f^{-1}\), is given by the formula: \[ \frac{d}{dx}[g(x)] = \frac{1}{f'(g(x))} \]. Here, \(g(x) = f^{-1}(x)\) and \(x = f(g(x))\).

Step 5: Evaluate \(g(0)\)

Since \(f(5) = 0\), \(f^{-1}(0) = 5\). Thus, \(g(0) = 5\).

Step 6: Compute \(f'(g(0))\)

Use \(g(0) = 5\) to find \(f'(5)\): \[f'(5) = 2(5) - 4 = 10 - 4 = 6\].

Step 7: Calculate \(d f^{-1} / dx\) at \(x=0\)

Substitute \(f'(5) = 6\) into the inverse derivative formula: \[ \frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))} \Rightarrow \frac{d}{dx}[f^{-1}(0)] = \frac{1}{f'(5)} = \frac{1}{6} \].

Step 8: Conclusion

The value of \( \frac{d}{dx}[f^{-1}(x)] \) at \(x = 0\) is \( \frac{1}{6} \).

Key Concepts

Inverse Functions

The concept of inverse functions is pivotal in math. An inverse function essentially reverses the effect of the original function. If you have a function, say \(f(x)\), that maps \(x\) to \(y\), then its inverse, \(f^{-1}(x)\), will map \(y\) back to \(x\). Inverse functions exist if a function is one-to-one. This means each output is only reached by one input. In simpler terms, it does not overlap itself.

• If \(y = f(x)\), then \(x = f^{-1}(y)\)
• The graph of an inverse function is a reflection over the line \(y = x\)

When dealing with quadratic functions like \(x^2 - 4x - 5\), finding an inverse can be tricky due to their possible non-one-to-one nature. This is why they may specify a domain, like \(x > 2\), ensuring that an inverse can exist.

Differentiation

Differentiation is a crucial technique in calculus used to find rates of change. It's essentially determining how a function changes as its input changes. For a simple function, \(f(x)\), the differential \(f'(x)\) gives the slope of the tangent line to the curve at any point on the graph.The rules of differentiation allow us to find derivatives quickly:

• The derivative of \(x^n\) is \(nx^{n-1}\)
• The derivative of constant terms are zero
• The sum of functions is the sum of their derivatives

In our exercise, differentiating \(f(x) = x^2 - 4x - 5\) follows these rules to produce \(f'(x) = 2x - 4\). This derivative helps us find the slope, which is crucial when applying the inverse function derivative formula.

Calculus

Calculus is the mathematical study of continuous change. Its two main branches are differentiation and integration. Differentiation, as we've discussed, involves finding how a function changes, while integration determines the area under curves. One interesting application of calculus is finding the derivative of inverse functions. The formula \[\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}\]allows us to work backward and determine how changes in \(y\) affect \(x\) in inverse scenarios.

• This formula tells us the derivative of the inverse function is one over the derivative of the direct function at the inverse point
• In the exercise, we used this to find \( \frac{1}{6} \), showing a gentle slope at the point

Calculus not only provides the tools to solve complex mathematical problems but also deepens our understanding of the relationships within different functions.