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Suppose that the differentiable function \(y=f(x)\) has an inverse and that the graph of \(f\) passes through the point (2,4) and has a slope of \(1 / 3\) there. Find the value of \(d f^{-1} / d x\) at \(x=4\)

Short Answer

Expert verified
The derivative \( \frac{d f^{-1}}{d x} \) at \( x=4 \) is 3.

Step by step solution

01

Understanding the Problem

We are given that the function \( y=f(x) \) has an inverse function, and we need to find the derivative of this inverse function, \( d f^{-1} / d x \), at \( x=4 \). We are also informed that \( f \) passes through the point \( (2,4) \) with a slope of \( 1/3 \).
02

Using the Inverse Derivative Formula

The derivative of the inverse function can be found using the formula: \[\frac{d f^{-1}}{d x} = \frac{1}{f'(f^{-1}(x))}.\] Here, \( x = 4 \) is a value of the inverse input.
03

Finding \( f^{-1}(4) \)

Since the function \( f \) passes through the point (2, 4), by definition of inverse functions, \( f^{-1}(4) = 2 \) because if \( f(2) = 4 \), then the inverse at 4 must return 2.
04

Finding \( f'(2) \)

We are given that the slope of \( f \) at \( x=2 \) is \( \frac{1}{3} \), which means \( f'(2) = \frac{1}{3} \).
05

Calculating \( \frac{d f^{-1}}{d x} \) at \( x=4 \)

Using our formula, \[\frac{d f^{-1}}{d x}\bigg|_{x=4} = \frac{1}{f'(f^{-1}(4))} = \frac{1}{f'(2)}.\] Substituting \( f'(2) = \frac{1}{3} \), we get \[\frac{d f^{-1}}{d x}\bigg|_{x=4} = \frac{1}{1/3} = 3.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiable Function
A differentiable function is one where the derivative exists at every point within its domain. The derivative represents the rate at which the function value is changing at any particular point, essentially capturing the function's slope. For a function to be differentiable at a point, it must be smooth and without any sharp turns or discontinuities.
This means that not only must the function be continuous—a function that you can draw without lifting your pencil—but it must have a well-defined tangent at every point. A classic example is a smooth curve rather than a corner or a cusp. For example:
  • If a curve has a cusp, like a sharp V-shape, it is not differentiable there.
  • If a curve has breaks or jumps, differentiability fails as well.
When solving problems involving differentiable functions, knowing whether a function is differentiable allows us to apply calculus tools like derivatives, making it a crucial aspect of analysis and a foundational element in understanding more complex mathematical concepts.
Inverse Functions
An inverse function essentially reverses what the original function does. If you have a function \( f(x) \), and it has an inverse, denoted \( f^{-1}(x) \), then \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \). This means that applying the function and its inverse to a value will bring you back to the original value.
Only one-to-one functions have inverses because each input to the function must correspond to exactly one output. Here's how you can determine if there's an inverse function:
  • The function must pass the horizontal line test. This test checks that any horizontal line only crosses the graph once.
  • Polynomials, trigonometric, and exponential functions often have inverses that help solve real-world problems by moving from effects back to causes.
An understanding of inverse functions is crucial for calculus, allowing us to find and work with derivative relationships in cases where direct calculation is complex.
Derivative Formula
The derivative formula for an inverse function is a neat tool that relates the derivatives of a function and its inverse. Given a differentiable function \( y=f(x) \) with an inverse, the formula is:
\[\frac{d f^{-1}}{d x} = \frac{1}{f'(f^{-1}(x))}.\]
This tells us that to compute the derivative of an inverse function at a point \( x \), you need to find where this input \( x \) maps in the original function, find the derivative of the original function there, and then take the reciprocal.
It's a powerful method because it reduces the problem to finding derivatives for functions whose forms might not be easily manipulable otherwise. Key points include:
  • Inverse derivative calculations often require understanding where the inverses meet their original.
  • This method is prominently used in calculus to handle complex mappings and transformations.
Mastering this formula is crucial for solving problems involving inversely related variables, often found in engineering and physics.

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Most popular questions from this chapter

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