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

1. Oct 4, 2015

### lep11

1. The problem statement, all variables and given/known data
Let f:[0,2] --> ]0,4[ be continuous function. Prove that ∃ a∈]0,2[ such that f(a)=a2

2. Relevant equations

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

2. Oct 4, 2015

### 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)?

3. Oct 4, 2015

### Krylov

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

4. Oct 6, 2015

### lep11

I still don't get it :(

5. Oct 6, 2015

### Krylov

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.)

6. Oct 6, 2015

### HallsofIvy

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

7. Oct 6, 2015

### 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

Last edited: Oct 6, 2015
8. Oct 6, 2015

### HallsofIvy

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

9. Oct 6, 2015

### Krylov

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 "$f(a) > u > f(b)$".

10. Oct 6, 2015

### lep11

....

Last edited: Oct 6, 2015
11. Oct 6, 2015

### 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 $g(2)$ ?

Last edited: Oct 6, 2015
12. Oct 6, 2015

### lep11

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!

13. Oct 6, 2015