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Chapter 3: Problem 48

Derivatives and inverse functions Find the slope of the curve \(y=f^{-1}(x)\) at (4,7) if the slope of the curve \(y=f(x)\) at (7,4) is \(\frac{2}{3}\)

Short Answer

Expert verified
Answer: The slope of the curve y=f^(-1)(x) at the point (4, 7) is 3/2.

Step by step solution

01

Identify given information

We are given that \(f(7) = 4\) and \(f'(7) = \frac{2}{3}\), and we need to find the slope of \(y = f^{-1}(x)\) at the point (4, 7).
02

Apply the inverse function formula

Using the formula \((f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\) and replacing x with 4, we have: \((f^{-1})'(4) = \frac{1}{f'(f^{-1}(4))}\) Since \(f^{-1}(4)=7\) (given that f(7)=4), the above equation becomes: \((f^{-1})'(4) = \frac{1}{f'(7)}\)
03

Plug in the given derivative of f(x) at x=7

We are given that \(f'(7) = \frac{2}{3}\). Therefore, we can substitute this value into the equation: \((f^{-1})'(4) = \frac{1}{\frac{2}{3}}\)
04

Calculate the slope of the inverse function

Now we invert the fraction inside the above equation: \((f^{-1})'(4) = \frac{1}{\frac{2}{3}} = \frac{1}{1} \cdot \frac{3}{2} = \frac{3}{2}\) So the slope of the curve \(y=f^{-1}(x)\) at the point (4, 7) is \(\frac{3}{2}\).

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Key Concepts

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

Derivatives
Derivatives are an essential tool in calculus that help us understand how a function changes at any given point. They can be viewed as the slope of a tangent line to a curve at a specific point.
For a function \( f(x) \), the derivative, denoted as \( f'(x) \), represents how \( y = f(x) \) changes as \( x \) changes. Derivatives are crucial for determining the rate of change of a function.
They can be applied in various subjects like physics, economics, and biology to model change and optimize results.
  • They measure the sensitivity of one variable concerning another.
  • Provide a concise way of understanding complex curves and shapes.
Understanding derivatives allows us to find and interpret the slope of a function, which can lead to finding maxima, minima, and understanding growth or decline over an interval.
Slope of a Curve
The slope of a curve at a particular point is determined using its derivative. This slope gives us the 'steepness' or inclination at that point.
For functions, we often talk about slopes in terms of the tangent line, which just touches the curve without intersecting it. If \( y = f(x) \), then the slope at any point \( x \) is given by \( f'(x) \).
Knowing the slope at a point can tell us a lot about the behavior of the function.
  • Positive slope: the function is increasing.
  • Negative slope: the function is decreasing.
  • Zero slope: indicates a possible minimum or maximum, often termed a critical point.
Slopes play a huge role when dealing with inverse functions, as understanding the original and the inverse curve's behavior at certain points helps us solve problems like the one in our exercise.
Inverse Function Theorem
The Inverse Function Theorem helps us find the derivative of an inverse function. It offers a way to correlate the derivative of a function with its inverse.
Specifically, if \( f(x) \) is invertible and differentiable, and its derivative at \( x \) is not zero, then the inverse \( f^{-1} \) is also differentiable at that corresponding point.
The formula provided by this theorem is:
  • \( \( (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} \) \)
This formula is what allows us to compute the slope of the inverse function, effectively translating changes in the original function into insights about the inverse.
This theorem is crucial whenever you deal with functions and their inverses, helping to bridge behaviors across corresponding points on both functions.

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Most popular questions from this chapter

Proof by induction: derivative of \(e^{k x}\) for positive integers \(k\) Proof by induction is a method in which one begins by showing that a statement, which involves positive integers, is true for a particular value (usually \(k=1\) ). In the second step, the statement is assumed to be true for \(k=n\), and the statement is proved for \(k=n+1,\) which concludes the proof. a. Show that \(\frac{d}{d x}\left(e^{k x}\right)=k e^{k x}\) for \(k=1\) b. Assume the rule is true for \(k=n\) (that is, assume \(\left.\frac{d}{d x}\left(e^{n x}\right)=n e^{n x}\right),\) and show this implies that the rule is true for \(k=n+1 .\) (Hint: Write \(e^{(n+1) x}\) as the product of two functions, and use the Product Rule.)

Proof of \(\frac{d}{d x}(\cos x)=-\sin x\) Use the definition of the derivative and the trigonometric identity $$ \cos (x+h)=\cos x \cos h-\sin x \sin h $$ to prove that \(\frac{d}{d x}(\cos x)=-\sin x\)

Vertical tangent lines a. Determine the points at which the curve \(x+y^{3}-y=1\) has a vertical tangent line (see Exercise 52 ). b. Does the curve have any horizontal tangent lines? Explain.

Consider the following functions (on the given interval, if specified). Find the inverse function, express it as a function of \(x,\) and find the derivative of the inverse function. $$f(x)=x^{2}-4, \text { for } x>0$$

An observer stands \(300 \mathrm{ft}\) from the launch site of a hot-air balloon. The balloon is launched vertically and maintains a constant upward velocity of \(20 \mathrm{ft} / \mathrm{s}\). What is the rate of change of the angle of elevation of the balloon when it is 400 ft from the ground? The angle of elevation is the angle \(\theta\) between the observer's line of sight to the balloon and the ground.
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