A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
The problem to use the intermediate value theorem to show that there is a root of the given equation in the specified interval.
sin(x)= x^2x, (1,2)
Can someone see if this is correct?
f(x) = sin(x)  x^2 + x, then f is continuous since sin(x) and x^2  x are continuous and their composition is also continuous. sin(x) = x^2  x which is equivalent to showing that f(x) = 0. f(1) = 0.841 and f(2) = 1.091, 0 s lying between f(1) and f(2). Since f is continuous, there is a c in (1,2) such that f(c) = 0
there is a root to the equation
sin(x) = x^2  x
anonymous
 3 years ago
The problem to use the intermediate value theorem to show that there is a root of the given equation in the specified interval. sin(x)= x^2x, (1,2) Can someone see if this is correct? f(x) = sin(x)  x^2 + x, then f is continuous since sin(x) and x^2  x are continuous and their composition is also continuous. sin(x) = x^2  x which is equivalent to showing that f(x) = 0. f(1) = 0.841 and f(2) = 1.091, 0 s lying between f(1) and f(2). Since f is continuous, there is a c in (1,2) such that f(c) = 0 there is a root to the equation sin(x) = x^2  x

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 can you check my work for me? :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm sure it is right, but I'm not this jim.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0LOL! okay :) is there a specific format the answer needs to be in?

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0I wouldn't use 'their composition'

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0'their difference' would be better

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0You are correct, all polynomials are continuous Sine is continuous, so sin(x)  x^2 + x is continuous f(1) = 0.841 and f(2) = 1.091 So because of the sign change and because f(x) is continuous, there is a number c such that f(c) = 0 since f(1) < 0 < f(2)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.