Prove that ∃ a∈]0,2[ such that f(a)=a^2

[lep11]

Homework Statement

Let f:[0,2] --> ]0,4[ be continuous function. Prove that ∃ a∈]0,2[ such that f(a)=a²

Homework Equations

The Attempt at a Solution

Sadly I have no idea whatsoever.
if e.g. a=1 and f(x)=x, then f(1)=1²

 

[HallsofIvy]

Look at the function g(x)= f(x)- x^2. g(0)= f(0) and g(2)= f(2)- 4. Since the largest possible value for f is 4, what can you say about g(2)?

 

[S.G. Janssens]

Apply the intermediate value theorem to a suitably chosen function ##g: [0,2] \to [0,4]##.

 

[lep11]

I still don't get it :(

 

[S.G. Janssens]

Is there something specific that you don't get? Did you make a start?

Could you perhaps give a precise formulate of the intermediate value theorem, as you understand it? (I suppose this question comes from a calculus course, where this theorem should appear.)

 

[HallsofIvy]

You were told to look at the function [itex]g(x)= f(x)- x^2[/itex]. Now you have been told to use the "intermediate value theorem". Do you know what that is?

 

[lep11]

I don't know how to get started and how to apply the intermediate value theorem.

Intermediate value theorem: If function f is continuous in [a,b] and f(a)<u<f(b), then there exists c ∈]a,b[ such that f(c)=u.

f(0)< a² <f(2) --> a∈]0,2[ such that f(a)=a²

 

[HallsofIvy]

And what have I already told you, last Sunday, about g(0) and g(4)?

 

[S.G. Janssens]

Careful, the function ##f## in your statement of the IVT is not the same as the function ##f## from your exercise. Replace ##f## in the IVT by ##g## as proposed earlier. Also, the IVT is true as well when we replace "##f(a) < u < f(b)##" by "##f(a) > u > f(b)##".

 

[lep11]

...

 

[geoffrey159]

You want to find a zero of the function ##g(x) = f(x) - x^2 ##.
If you prove there is ##x_0## and ##x_1## in ##[0,2]## such that ##g(x_0) < 0 ## and ##g(x_1) > 0 ##, then the intermediate value theorem guarantees the existence of a zero for ##g##.

You have ## 0 < f < 4 ##. Then what about ##g(0)## and ##g(2)## ?

 

[lep11]

You want to find a zero of the function ##g(x) = f(x) - x^2 ##.
If you prove there is ##x_0## and ##x_1## in ##[0,2]## such that ##g(x_0) < 0 ## and ##g(x_1) > 0 ##, then the intermediate value theorem guarantees the existence of a zero for ##g##.

You have ## 0 < f < 4 ##. Then what about ##g(0)## and ##g(2)## ?

g(0)>0 and g(2)<0
g(0)>0>g(2) so IVT: ∃x₀∈]0,2[ such that g(x₀)=0 ⇔ f(x₀)=x₀²
Now I got it, I appreciate your help! thanks everyone!

 

[geoffrey159]

g(0)>0 and g(2)<0
g(0)>0>g(2) and g continuous on [0,2] so IVT: ∃x₀∈]0,2[ such that g(x₀)=0 ⇔ f(x₀)=x₀²
Now I got it, I appreciate your help! thanks everyone!