How Does Sketching Graphs Help Find the Real Root of f(x) = 2sin(x) - 3x + 2?

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Homework Help Overview

The discussion revolves around the function f(x) = 2sin(x) - 3x + 2 and the task of sketching graphs to identify its real root. Participants are exploring how to visualize the function alongside a linear equation to find points of intersection.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants suggest sketching the graphs of y = 2sin(x) and y = 3x - 2 to find intersections. There is also a discussion about terminology, questioning the use of "roots" versus "zeroes." Some participants express confusion about the implications of the problem statement and the nature of the functions involved.

Discussion Status

The conversation is ongoing, with participants sharing their attempts to graph the functions and analyze their intersections. One participant notes a potential issue with their calculations related to the Newton-Raphson method, indicating a lack of consensus on the approach but a willingness to explore the problem further.

Contextual Notes

There is mention of a calculator mode affecting calculations, which highlights potential constraints in the problem-solving process. Additionally, the original poster's confusion about the function's characteristics suggests that assumptions about the problem may need clarification.

Bucky
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"By sketching appropriate graphs (one of them a straight line) show that the function
f(x) = 2sin(x) - 3x + 2

has only one real root."

this one confuses me from the word go. how can you sketch a straight line of that function when it has sin as part of it?
 
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The problem did say "graphs" (plural).
How about y= 2 sin(x) and y= 3x- 2?

Where the graphs cross, 2 sin(x)= 3x- 2 so 3 sin(x)- 3x+ 2= 0.

(In the interest of beating a dead horse)
Was that the exact wording of the problem? A function does not have "roots". A function has "zeroes", which are "roots" of the equation f(x)= 0.

If I saw something like "find the roots of f(x)= x2" I would think "x2= 1 or x2= 4 or x2= 1000...?"
 
Last edited by a moderator:
yeah that's a word for word quote.

ok so i set the two functions up and compared them and they cross once at approx 1.3, 1.9

i thought that the 1.3 was the value that has to be used in the Newton rhapson method, but the first iteration derivates so heavily from the origonal (i ended up with -0.49908) that i think I am wrong.

EDIT: Yeah somethings wrong...the more iterations i do the more the number jumps around.

DOUBLE EDIT: Calculator was in wrong mode, I am an idiot.
 
Last edited:
DOUBLE EDIT: Calculator was in wrong mode, I am an idiot.
Well, I wasn't going to mention it---
 

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