# Find The Range of Function.

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.

## Answers and Replies

tiny-tim
Science Advisor
Homework Helper
hi klen! (try using the X2 button just above the Reply box )
(e^-x)f(x) ≥ (x^2)/2 + f'(0)x.

or you could say let g(x) = e-xf(x) …

then g''(x) ≥ 1 (btw, how did you get on with your series question?)

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.

tiny-tim
Science Advisor
Homework Helper
excellent! (good LaTeX too )

in each of these sums the first two are negative terms and last two are positive,
so the sum evaluates to,
$$S_n = \sum_{k=0}^{n} {(32k-12)}$$

How do you get to this step? 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,
$$S_n= \sum_{i=1}^n {(32i-12)}$$

• 1 person
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,
$$S_n= \sum_{i=1}^n {(32i-12)}$$

That was cool! Thanks! 