Let f(x)=20/(x^6+x^4+x^2+1). I need to show that for any k in [5,20], there is a point c in [0,1] such that f(c)=k.

rivasguss9 2022-08-07 Answered
Let f ( x ) = 20 ( x 6 + x 4 + x 2 + 1 ) .
I need to show that for any k [ 5 , 20 ], there is a point c [ 0 , 1 ] such that f ( c ) = k.
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)

prelatiuvq
Answered 2022-08-08 Author has 6 answers
Observe that x 6 + x 4 + x 2 + 1 > 0 x R and is continuous. Now f ( x ) is continuous on R and observe that f ( 1 ) = 5 , f ( 0 ) = 20. Intermediate value theorem tells that for every k such that 5 k 20 there exists a c [ 0 , 1 ] such that f ( c ) = k.
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

You might be interested in

asked 2022-08-19
f ( x ) = 1 1 | x | + 1 x 4
I was asked to show though the intermediate value theorem that f ( x ) has at least one solution when f ( x ) = 314. I found that f ( x ) is continuous when 1 < x < 0 and 0 < x < 1. The problem I'm having is that I thought that the theorem only worked for closed intervalls. Any tips are greatly appreciated!
asked 2022-06-28
The Intermediate Value Theorem has been proved already: a continuous function on an interval [ a , b ] attains all values between f ( a ) and f ( b ). Now I have this problem:

Verify the Intermediate Value Theorem if f ( x ) = x + 1 in the interval is [ 8 , 35 ].

I know that the given function is continuous throughout that interval. But, mathematically, I do not know how to verify the theorem. What should be done here?
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 2022-08-09
Suppose f : [ 1 , 1 ] R is continuous, f ( 1 ) > 1, and f ( 1 ) < 1. Show that f has a fixed point.
So for every y between f ( 1 ) and f ( 1 ) we can find an x s.t. 1 < x < 1 and f ( x ) = y, but I'm not sure how to show one of these creates a fixed point.
asked 2022-05-26
Prove that f ( x ) = 3 has a solution on the interval [ a , b ]

And Intermediate Value Theorem says that

if f ( a ) f ( b ) < 0, then it has a solution on that interval

So instead of seeing if 3 is between the interval and stuff like that. Can't I just do this:
f ( x ) = 3
f ( x ) 3 = 0
And then we consider f ( x ) 3 a completely new whole function called g ( x ) = 0

According to the theorem, I can say that since g ( a ) g ( b ) < 0, it has a solution on the interval.
asked 2022-08-11
Given f : [ 0 , 1 ] R 2 such that f ( 0 ) = ( 1 , 0 ) , f ( 1 ) = ( 1 , 0 ) and f is continuous with respect to the standard topologies. How can I prove that there exists x [ 0 , 1 ] such that f ( x ) = ( 0 , y ), for some y R ?
My attempt to a solution was to consider the restriction of f to the set
A = { x : f ( x ) = ( y , 0 )   for some   y R } [ 0 , 1 ] .
Since f is continuous then its restriction is continuous.
Since the restriction of f to A is "basically" a map from A to R , it makes sense that I would just need to apply IVT to finish the proof. I'm having issue in formalizing the reasoning that the restriction is "basically" a map from A to R .
asked 2022-07-10
The Intermediate Value Theorem states that, for a continuous function f : [ a , b ] R , if f ( a ) < d < f ( b ), then there exists a c ( a , b ) such that f ( c ) = d .

I wonder if I change the hypothesis of f ( a ) < d < f ( b ) to f ( a ) > d > f ( b ), the result still holds. I believe so, since f assumes a fixed point f ( x ) = x in [ a , b ], so we would have c = d, although I'm not completely sure.

I need this result in order to prove the set X = { x [ a , b ] s . t . f | [ a , x ] is bounded }, with a continuous f, is not empty.

New questions