/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 72 Suppose \(y=L(x)=a x+b\) (with \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 72

Suppose \(y=L(x)=a x+b\) (with \(a \neq 0\) ) is the equation of the line tangent to the graph of a one-to-one function \(f\) at \(\left(x_{0}, y_{0}\right) .\) Also, suppose that \(y=M(x)=c x+d\) is the equation of the line tangent to the graph of \(f^{-1}\) at \(\left(y_{0}, x_{0}\right)\) a. Express \(a\) and \(b\) in terms of \(x_{0}\) and \(y_{0}\) b. Express \(c\) in terms of \(a,\) and \(d\) in terms of \(a, x_{0},\) and \(y_{0}\) c. Prove that \(L^{-1}(x)=M(x)\)

Short Answer

Expert verified
Question: Prove that if L(x) = ax + b is tangent to the graph of a one-to-one function at (x0, y0) and M(x) = cx + d is tangent to the graph of its inverse at (y0, x0), then L^(-1)(x) = M(x).

Step by step solution

01

We are given that y = L(x) = ax + b is tangent to the graph of f at (x0, y0) and y = M(x) = cx + d is tangent to the graph of f^(-1) at (y0, x0). Step 2: Express a and b

We know that the tangent L(x) passes through the point (x0, y0). So, we can plug in the coordinates into L(x) to find a and b in terms of x0 and y0: y0 = ax0 + b Now, we rewrite above equation as b = y0 - ax0. So, we have expressed a and b in terms of x0 and y0 as follows: a = a b = y0 - ax0 #b. Express c in terms of a, and d in terms of a, x0, and y0# Step 3: Derivatives of functions
02

In order to find c, we have to use the fact that if \((f^{-1})'(y_0)= \frac{1}{f'(x_0)}\). Since L(x) is tangent to the graph of f at (x0, y0), its slope must be equal to the derivative of f evaluated at x0: \(f'(x_0)=a\) For function M(x), its slope must be the derivative of the inverse of f evaluated at y0. Therefore, we have: \((f^{-1})'(y_0)=c\) Step 4: Relating c to a

Using the relationship from Step 3: \(c=(f^{-1})'(y_0)= \frac{1}{f'(x_0)}= \frac{1}{a}\) Step 5: Finding d
03

Since M(x) passes through the point (y0, x0), we can plug in the coordinates into M(x) to find d in terms of c, x0, and y0: \(x_0=cy_0+d\) Substituting c: \(x_0=\frac{1}{a}y_0+d\) Now, we rewrite above equation as d = x0 - y0/a So, we have expressed c and d in terms of a, x0, and y0 as follows: c = 1/a d = x0 - y0/a #c. Prove that L^(-1)(x) = M(x)# Step 6: Find inverse of L(x)

We are given L(x) = ax + b, substitute b from Step 2: L(x) = ax + y0 - ax0 Now, find the inverse function: x = ay + y0 - ay0 x - y0 + ay0 = ay y = (x - y0 + ay0)/a This is the inverse of L(x): L^(-1)(x) = (x - y0 + ay0)/a Step 7: Compare L^(-1)(x) with M(x)
04

We have M(x) = cx + d from above. Substitute c and d from Step 5: M(x) = (1/a)x + x0 - y0/a Now, rearrange M(x): M(x) = (x - y0 + ay0)/a Step 8: Conclusion

We can see that L^(-1)(x) = M(x), which concludes the proof.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Witch of Agnesi Let \(y\left(x^{2}+4\right)=8\) (see figure). a. Use implicit differentiation to find \(\frac{d y}{d x}\) b. Find equations of all lines tangent to the curve \(y\left(x^{2}+4\right)=8\) when \(y=1\) c. Solve the equation \(y\left(x^{2}+4\right)=8\) for \(y\) to find an explicit expression for \(y\) and then calculate \(\frac{d y}{d x}\) d. Verify that the results of parts (a) and (c) are consistent.

Prove that \(\frac{d^{2 n}}{d x^{2 n}}(\sin x)=(-1)^{n} \sin x\) and \(\frac{d^{2 n}}{d x^{2 n}}(\cos x)=(-1)^{n} \cos x\)

A port and a radar station are 2 mi apart on a straight shore running east and west. A ship leaves the port at noon traveling northeast at a rate of \(15 \mathrm{mi} / \mathrm{hr}\). If the ship maintains its speed and course, what is the rate of change of the tracking angle \(\theta\) between the shore and the line between the radar station and the ship at 12: 30 p.m.? (Hint: Use the Law of sines.)

The bottom of a large theater screen is \(3 \mathrm{ft}\) above your eye level and the top of the screen is \(10 \mathrm{ft}\) above your eye level. Assume you walk away from the screen (perpendicular to the screen) at a rate of \(3 \mathrm{ft} / \mathrm{s}\) while looking at the screen. What is the rate of change of the viewing angle \(\theta\) when you are \(30 \mathrm{ft}\) from the wall on which the screen hangs, assuming the floor is horizontal (see figure)?

Special Quotient Rule In general, the derivative of a quotient is not the quotient of the derivatives. Find nonconstant functions \(f\) and \(g\) such that the derivative of \(f / g\) equals \(f^{\prime} / g^{\prime}\).
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.