Question

Find the derivative of the inverse of the following finctions at the specified point on the graph of the inverse function. You do not need to find \(f^{-1}\). $$f(x)=x^{2}-2 x-3,\( for \)x \leq 1 ;(12,-3)$$

Short answer

Question: Find the derivative of the inverse function of \(f(x) = x^2 - 2x - 3\) for \(x\leq1\) at the point \((12, -3)\) on the graph of the inverse function. Answer: The derivative of the inverse function, \(f^{-1}(x)\), at the point \((12,-3)\) is \(\frac{1}{-2}\).

Step-by-step solution

Write the function and the point on the inverse function graph

We have the function: $$f(x) = x^2 - 2x - 3 \quad\text{for} \quad x\leq1$$ The point on the graph of the inverse function is: $$(12, -3)$$

Use the property of the inverse function derivative

By the property of the inverse function, we have: $$\left(\frac{d}{dx}f^{-1}(x)\right) = \frac{1}{f'(f^{-1}(x))}$$ For this, we need to find the derivative of the original function \(f(x)\) first.

Find the derivative of f(x) with respect to x

Using the power rule, find the derivative of the given function: $$f'(x) = \frac{d}{dx} (x^2 - 2x - 3)$$ $$f'(x) = 2x - 2$$

Find the relation between x and y on the graph of the inverse function

We know that the point \((12, -3)\) is on the graph of the inverse function \(f^{-1}(x)\). So, we have the following relationship: $$y = f^{-1}(12) \quad \text{and} \quad x = f^{-1}(-3)$$ Since \((x,y)\) is a point on the graph of \(f\), we also have: $$y = f(x)$$

Find the x-coordinate corresponding to the point on the graph of the original function

Using the relationship, we found in Step 4: $$-3 = f(x) = x^2 - 2x - 3$$ Solving for x, we get: $$0 = x^2 - 2x$$ Since we know that \(x \leq 1\), the solution for x is: $$x = 0$$ Now, we have the point \((0, -3)\) on the graph of the original function \(f(x)\).

Calculate f'(0) and use the property of the inverse function derivative

From Step 3, the derivative of the function is: $$f'(x) = 2x - 2$$ So, \(f'(0) = 2\cdot 0 - 2 = -2\). Now, using the property of the inverse function derivative we derived in Step 2: $$\left(\frac{d}{dx}f^{-1}(x)\right) = \frac{1}{f'(f^{-1}(x))}$$ Plugging in the values: $$\left(\frac{d}{dx}f^{-1}(12)\right) = \frac{1}{f'(0)} = \frac{1}{-2}$$ So, the derivative of the inverse function, \(f^{-1}(x)\), at the specified point \((12,-3)\) is: $$\frac{d}{dx}f^{-1}(12) = \frac{1}{-2}$$

Key concepts

Derivative of Inverse Function

Understanding the derivative of an inverse function is crucial in calculus, particularly when we need to find the slope of a curve at a given point without explicitly knowing the inverse function. Given a function, say, \( f \), if it has an inverse \( f^{-1} \), and both are differentiable, then we can relate their derivatives using the following key relationship:
\[ \left(\frac{d}{dx} f^{-1}(x)\right) = \frac{1}{f'(f^{-1}(x))} \]
This powerful formula allows us to calculate the derivative of \( f^{-1} \) at a certain point by merely knowing the derivative of \( f \) at the corresponding point. It's essential to note that this method hinges on the point \( (x, y) \) on the graph of \( f \) being related to a certain point on the graph of its inverse, \( f^{-1} \), thus enabling us to use the relationship between \( x \) and \( y \) to find derivative values without computing the inverse explicitly.

Calculus

Calculus is a vast field that provides tools for understanding change and motion. It's divided into two main branches: differential calculus and integral calculus. Differential calculus, which is of primary concern here, involves the study of rates at which quantities change, a concept encapsulated by the notion of a derivative. The derivative measures how a function's output value changes concerning changes in the input value. This is particularly useful for finding tangents to curves, optimizing functions, and understanding motion over time. Calculus operates on the principles of limits, which allow mathematicians to explore values at infinitesimally close points, approaching accuracy that is otherwise unreachable through basic algebraic methods.

Power Rule

The power rule is a quick and straightforward rule for differentiating functions of the form \( f(x) = x^n \), where \( n \) is a real number. According to the power rule:
\[ f'(x) = nx^{n-1} \]
This rule dramatically simplifies the process of differentiation because we can directly apply it to find the derivative of each term in a polynomial. For example, to differentiate \( f(x) = x^2 - 2x - 3 \) concerning \( x \), we apply the power rule term-by-term, giving \( f'(x) = 2x - 2 \). It's a cornerstone of differential calculus and is regularly used in tandem with other differentiation techniques to tackle more complex functions.

Differentiation

Differentiation is a fundamental process in calculus that focuses on finding the derivative of a function, denoted as \( f'(x) \) or \( \frac{df}{dx} \). It is the action of computing a derivative, which is the measure of how a function value changes as its input changes. The derivative itself can be a function that provides the slopes of the tangent lines to the curve represented by the original function at every point. Differentiation involves rules and procedures like the power rule, the product rule, the quotient rule, and the chain rule to tackle various functions. Through differentiation, we can solve problems related to rates of change in physics, economics, biology, and virtually every field that involves quantitative analysis.