What is the Range of a Twice Differentiable Function?

In summary, the function f is twice differentiable and has f(0)=f(1)=0. For 0<x<1, the following is true: 0<f(x)<∞, -1/2<f(x)<1/2, -1/4<f(x)<1, -∞<f(x)<0.
  • #1
klen
41
1
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.
 
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  • #2
hi klen! :smile:

(try using the X2 button just above the Reply box :wink:)
klen said:
(e^-x)f(x) ≥ (x^2)/2 + f'(0)x.

or you could say let g(x) = e-xf(x) …

then g''(x) ≥ 1 :wink:

(btw, how did you get on with your series question?)
 
  • #3
Hi Tiny Tim,

Thanks for the help.

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

(good LaTeX too :wink:)
 
  • #5
klen said:
in each of these sums the first two are negative terms and last two are positive,
so the sum evaluates to,
[tex] S_n = \sum_{k=0}^{n} {(32k-12)}[/tex]

How do you get to this step? :confused:
 
  • #6
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,
[tex]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}[/tex]
which after solving comes,
[tex]S_n= \sum_{i=1}^n {(32i-12)}[/tex]
 
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  • #7
klen said:
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,
[tex]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}[/tex]
which after solving comes,
[tex]S_n= \sum_{i=1}^n {(32i-12)}[/tex]

That was cool! :cool:

Thanks! :smile:
 

FAQ: What is the Range of a Twice Differentiable Function?

1. What is the range of a function?

The range of a function is the set of all possible output values that the function can produce.

2. How do you find the range of a function?

To find the range of a function, you need to first determine all the possible input values (domain) and then compute the corresponding output values. The range will be the set of all these output values.

3. Can the range of a function be infinite?

Yes, the range of a function can be infinite if the function has an unbounded domain. For example, the function f(x) = x² has a range of all positive real numbers, which is infinite.

4. What does the range represent in a graph of a function?

In a graph of a function, the range represents the vertical axis (y-axis) and it shows all the possible y-values that the function can produce for different input values (x-values).

5. Why is it important to find the range of a function?

Finding the range of a function is important because it helps us understand the behavior of the function. It also helps us determine if a function is one-to-one (injective) or onto (surjective), which has important applications in various fields of mathematics and science.

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