# Find The Range of Function.

1. Jun 6, 2013

### klen

A function f :[0,1]→ℝ is twice differentiable with f(0)=f(1)=0 such that

f''(x) - 2f'(x) + f(x) ≥ e^x, x ε [0,1].

Then for 0<x<1 which of the following is true
0<f(x)<∞, -1/2<f(x)<1/2, -1/4<f(x)<1, -∞<f(x)<0.

I tried to solve it in the following way:
I tried to take the e^x to the left side and found that the equation becomes,
[(e^-x)f'(x) - (e^-x)f(x)]' ≥ 1, thus on integration,

(e^-x)f'(x) - (e^-x)f(x) -f'(0) ≥ x which is same as

[(e^-x)f(x)]' ≥ x on integration

(e^-x)f(x) ≥ (x^2)/2 + f'(0)x.
I don't know how to proceed from here.

2. Jun 6, 2013

### tiny-tim

hi klen!

(try using the X2 button just above the Reply box )
or you could say let g(x) = e-xf(x) …

then g''(x) ≥ 1

(btw, how did you get on with your series question?)

3. Jun 7, 2013

### klen

Hi Tiny Tim,

Thanks for the help.

in the series question, we can break the sum as groups of four, like this:
$$S_n = \sum_{k=1}^{4} {(-1)^{k(k+1)/2}}{k^2} + \sum_{k=5}^{8} {(-1)^{k(k+1)/2}}{k^2} + ...$$
and so on,
in each of these sums the first two are negative terms and last two are positive,
so the sum evaluates to,
$$S_n = \sum_{k=0}^{n} {(32k-12)}$$
this gives the sum to be
$$S_n= {16n^2+4n}$$
which is possible only for first and fourth values.

4. Jun 7, 2013

### tiny-tim

excellent!

(good LaTeX too )

5. Jun 7, 2013

### Saitama

How do you get to this step?

6. Jun 7, 2013

### klen

In these sums for the first terms we have k=1,5,9...
so k is of form k=4i-3, i=1,2,3....
similarly for the next successive terms k is of form 4i-2, 4i-1, 4i.
Since the first two terms are positive and last two are negative we have,
$$S_n= \sum_{i=1}^n {-(4i-3)^2} + \sum_{i=1}^n {-(4i-2)^2} + \sum_{i=1}^n {(4i-1)^2} + \sum_{i=1}^n {(4i)^2}$$
which after solving comes,
$$S_n= \sum_{i=1}^n {(32i-12)}$$

7. Jun 7, 2013

### Saitama

That was cool!

Thanks!