This is a 2 part question. Please assist with both images. Thanks so much **Problem Statement:**Given the equation $(x-2)^2 - \ln x = 0$.**Objective:**Show that this equation has a solution on the interval $[e, 4]$.**Explanation:**To solve this problem, you might consider using methods like the Intermediate Value Theorem. This theorem suggests that if a continuous function changes sign over an interval, it must have a root in that interval. Here’s a brief guide on how you might approach this:1. **Function Definition:** Define the function $ f(x) = (x-2)^2 - \ln x $.2. **Continuity:** Check if $ f(x) $ is continuous on $[e, 4]$. Since polynomials and the natural logarithm ($\ln x$) are both continuous where defined, and their sum/difference is also continuous, $ f(x) $ is continuous on this interval.3. **Sign Change:** Evaluate $ f(e) $ and $ f(4) $: - Calculate $ f(e) = (e-2)^2 - \ln e = (e-2)^2 - 1 $. - Calculate $ f(4) = (4-2)^2 - \ln 4 = 4 - \ln 4 $. Determine if $ f(e) $ and $ f(4) $ have opposite signs. If they do, by the Intermediate Value Theorem, there is at least one $ c $ in the interval $[e, 4]$ such that $ f(c) = 0 $.4. **Conclusion:** If a sign change is confirmed, then the solution exists within the specified interval.By verifying these steps, you can show that the given equation has a solution on $[e, 4]$. **Problem Statement:**Given $ f(x) = (x - 2)^2 - \ln x = 0 $. Show that it has a solution on the interval $[1, 2]$.**Explanation:**The problem asks to demonstrate that the equation $ f(x) = (x - 2)^2 - \ln x = 0 $ has at least one solution within the interval $[1, 2]$.**Approach:**1. **Substitute Endpoints:** - Calculate $ f(1) $ and $ f(2) $ to find the values of the function at the endpoints of the interval. 2. **Continuous Function:** - Show that $ f(x) $ is continuous on $[1, 2]$. The function $ (x - 2)^2 $ is a polynomial, hence continuous everywhere, and $-\ln x$ is continuous for $ x > 0 $.3. **Intermediate Value Theorem (IVT):** - Apply the Intermediate Value Theorem, which states that if a function $ f $ is continuous on a closed interval $[a, b]$ and $ f(a) $ and $ f(b) $ have opposite signs, then there exists at least one $ c $ in $(a, b)$ such that $ f(c) = 0 $.By following these steps, you can show the existence of a solution for the equation within the given interval.

Answered: Image /qna-images/answer/4e4987e8-ed3f-4a46-8bf9-ce85f2a24b80.jpg

Answered: Given (x-2)²-Inx=0. Show that it has a…

Given (x-2)²-Inx=0. Show that it has a solution on [e, 4].

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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