suppose we have following function f <mo stretchy="false">( x <mo stretchy="false">)

veleumnihryz 2022-05-02 Answered
suppose we have following function
f ( x ) = x 2 + 10 sin ( x )
we should show that there exist number c such that f ( c ) = 1000

clearly we can solve this problem using intermediate value theorem for instance
f ( 0 ) = 0
f ( 90 ) = 90 2 + 10 sin ( 90 ) = 8100 + 10 1 = 8110
and because 1000 is between these two number we can see that there exist such c so that f ( c ) = 1000

am i right? thanks in advance
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Trey Harrington
Answered 2022-05-03 Author has 10 answers
f ( x ) = 2 x + 10 cos x is clearly positive if x 6, hence our function is increasing over the interval [ 6 , 33 ] and since 1000 is between f ( 6 ) and f ( 33 ), there is some (only one) c ( 6 , 33 ) such that f ( c ) = 1000.
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2022-05-08
I have been stuck on this Real Analysis problem for hours and am just totally clueless- I am sure it is some application of the Intermediate Value Theorem- suppose   f : R R is continuous at every point. Prove that the equation   f ( x ) = c cannot have exactly two solutions for every value of   c .

Would appreciate some help
asked 2022-04-07
Suppose that f : [ a , b ] R is continuous and that f ( a ) < 0 and f ( b ) > 0. By the intermediate-value theorem, the set S = { x [ a , b ] : f ( x ) = 0 } is nonempty. If c = sup S, prove that c S.

My first thought was to show that S is finite therefore c S, but there is no guarantee that f doesn't have infinitely many zeros.

A thought I have now is to show that c > max ( S ) can not be true, but I do not know how to show this, is this even the correct thing to show? Thank you in advance for any input.
asked 2022-05-08
Given equation x 100 + x 99 + x 98 + . . . + x + 1 = 5 I need to determine whether there exists one solution x such that x > 0. Also, whether I have one solution x < 0.

First, I let a = 0 such that f ( a ) = 1. Let b = 1 such that f ( b ) = 101. Since f ( a ) < 5 < f ( b ), thus, by intermediate value theorem, there exits one solution x ( 0 , 1 ). Similarly, I can conclude that there exits one solution smaller than 0. Is my process correct?
asked 2022-05-09
Suppose that f is a continuous function and that f ( 1 ) = f ( 1 ) = 0. Show that there is c ( 1 , 1 ) such that
f ( c ) = c 1 c 2
I am not sure if this is a new question as I set it this morning, after solved a similar question. I wanted to prove it using the same idea (Intermediate Value Theorem) but it was not that nice...

Can anyone help me on this?

By the way, the 'similar question' I mentioned above is as follow:

Suppose that f is a continuous function and that f ( 0 ) = 1 and f ( 1 ) = 2. Show that there is c ( 0 , 1 ) such that
f ( c ) = 1 c
The hint given was to let g ( x ) = x f ( x ) and use the Intermediate Value Theorem.
asked 2022-05-10
I am trying to understand and prove the fundamental theorem of calculus and I ran into some confusion understanding the intermediate value theorem . several sources online claim that if a function f(x) is continuous on [a,b] let s be a number such that f ( a ) < s < f ( b ) then there exists a number k in the open interval (a,b) such that f(k)=s my question is why do we only assume the open interval shouldn't it also include the closed interval [a,b] and also why does s have to be less than both f ( a ) and f ( b )?
asked 2022-04-30
Let I = [ a , b ] with a < b and let u : I R be a function with bounded pointwise variation, i.e.
V a r I u = sup { i = 1 n | u ( x i ) u ( x i 1 ) | } <
where the supremum is taken over all partition P = { a = x 0 < x 1 < . . . < x n 1 < b = x n }. How can I prove that if u satisfies the intermediate value theorem (IVT), then u is continuous?

My try: u can be written as a difference of two increasing functions f 1 , f 2 . I know that a increasing function that satisfies the (ITV) is continuous, hence, if I prove that f 1 , f 2 satisfies the (ITV) the assertion follows. But, is this true? I mean, f 1 , f 2 satisfies (ITV)?
asked 2022-05-09
Let a , b R , a < b and let f be a differentiable real-valued function on an open subset of R that contains [a,b]. Show that if γ is any real number between f ( a ) and f ( b ) then there exists a number c ( a , b ) such that γ = f ( c ).

Hint: Combine mean value theorem with the intermediate value theorem for the function ( f ( x 1 ) f ( x 2 ) ) x 1 x 2 on the set { ( x 1 , x 2 ) E 2 : a x 1 < x 2 b }.

I am having a lot of trouble trying to start on this problem.