URGENT GRE 9367 Q42 AND Q23

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 Joined: Tue Nov 01, 2011 5:44 am
URGENT GRE 9367 Q42 AND Q23
42. What is the greatest value of b for which any real valued function f that satisfies the following properties must also satisfy f(1)<5 ?
1> f is infinitely differentiable on the real numbers
2> f(0)=1, f'(0)=1, and f''(0)=2; and
3> f'''(x)< b for all x in [0,1]
a)1
b)2
c)6
d)12
e)24
23.Let f be a real valued function continuous on the close interval [0,1] and differentiable on the open interval (0,1) with f(0)=1 and f(1)=0. Which of the following must be true
1. there exists x in (0,1) such that f(x)=x
2. there exists x in (0,1) such that f'(x)=1
3. f(x)>0 for all x in [0,1)
a) 1 only
b) 2 only
c) 1 and 2 only
d) 2 and 3 only
e) 1.2 and 3
Please post as soon as possible. I am in a big mess regarding this questions. Please Help.
1> f is infinitely differentiable on the real numbers
2> f(0)=1, f'(0)=1, and f''(0)=2; and
3> f'''(x)< b for all x in [0,1]
a)1
b)2
c)6
d)12
e)24
23.Let f be a real valued function continuous on the close interval [0,1] and differentiable on the open interval (0,1) with f(0)=1 and f(1)=0. Which of the following must be true
1. there exists x in (0,1) such that f(x)=x
2. there exists x in (0,1) such that f'(x)=1
3. f(x)>0 for all x in [0,1)
a) 1 only
b) 2 only
c) 1 and 2 only
d) 2 and 3 only
e) 1.2 and 3
Please post as soon as possible. I am in a big mess regarding this questions. Please Help.
Re: URGENT GRE 9367 Q42 AND Q23
23.lifeisgood88 wrote:42. What is the greatest value of b for which any real valued function f that satisfies the following properties must also satisfy f(1)<5 ?
1> f is infinitely differentiable on the real numbers
2> f(0)=1, f'(0)=1, and f''(0)=2; and
3> f'''(x)< b for all x in [0,1]
a)1
b)2
c)6
d)12
e)24
23.Let f be a real valued function continuous on the close interval [0,1] and differentiable on the open interval (0,1) with f(0)=1 and f(1)=0. Which of the following must be true
1. there exists x in (0,1) such that f(x)=x
2. there exists x in (0,1) such that f'(x)=1
3. f(x)>0 for all x in [0,1)
a) 1 only
b) 2 only
c) 1 and 2 only
d) 2 and 3 only
e) 1.2 and 3
Please post as soon as possible. I am in a big mess regarding this questions. Please Help.
according to intermediate value theorem, 1 is obvious.
according to mean value theorem, 2 is obvious.
and the last one is still obvious.....show u a example:y=2x^23x+1
42.(I hope you know the language of Wolfram's softwareMathematica...)
Integrate[f'''(x),{x,0,1}]=f''(1)f''(0), and so do f''(x) and f'(x)
5>Max[f(1)]=Max[Integrate[f'(x),{x,0,1}]+f(0)=Max[Integrate[f''(x),{x,0,1}]+f'(0)+f(0)=Max[Integrate[f'''(x),{x,0,1}]+f''(0)+f'(0)+f(0)
Apparently, Max[f'''(x)]=b > Max[f''(x)]=bx+2 > Max[f'(x)]=(b/2)*x^2+2x+1 > Max[f(x)]= ((b^3)/6)x^3+x^2+x+1
So Max[f(1)]=b/6+1+1+1<5 > b<=12
I'm not a math major student. If you think my solution is not convinced, you can prove it yourself..=_=iiii

 Posts: 5
 Joined: Tue Nov 01, 2011 5:44 am
Re: URGENT GRE 9367 Q42 AND Q23
Thanks a lot... but can you explain me the intermediate value theorem part.... i am bit confused about it... as it states that if f(x) is continuous on [a,b] and k is any number between f(a) and f(b), then there exists at least one no. c such that a<c<b and f(c)=k.mhyyh wrote:23.lifeisgood88 wrote:42. What is the greatest value of b for which any real valued function f that satisfies the following properties must also satisfy f(1)<5 ?
1> f is infinitely differentiable on the real numbers
2> f(0)=1, f'(0)=1, and f''(0)=2; and
3> f'''(x)< b for all x in [0,1]
a)1
b)2
c)6
d)12
e)24
23.Let f be a real valued function continuous on the close interval [0,1] and differentiable on the open interval (0,1) with f(0)=1 and f(1)=0. Which of the following must be true
1. there exists x in (0,1) such that f(x)=x
2. there exists x in (0,1) such that f'(x)=1
3. f(x)>0 for all x in [0,1)
a) 1 only
b) 2 only
c) 1 and 2 only
d) 2 and 3 only
e) 1.2 and 3
Please post as soon as possible. I am in a big mess regarding this questions. Please Help.
according to intermediate value theorem, 1 is obvious.
according to mean value theorem, 2 is obvious.
and the last one is still obvious.....show u a example:y=2x^23x+1
42.(I hope you know the language of Wolfram's softwareMathematica...)
Integrate[f'''(x),{x,0,1}]=f''(1)f''(0), and so do f''(x) and f'(x)
5>Max[f(1)]=Max[Integrate[f'(x),{x,0,1}]+f(0)=Max[Integrate[f''(x),{x,0,1}]+f'(0)+f(0)=Max[Integrate[f'''(x),{x,0,1}]+f''(0)+f'(0)+f(0)
Apparently, Max[f'''(x)]=b > Max[f''(x)]=bx+2 > Max[f'(x)]=(b/2)*x^2+2x+1 > Max[f(x)]= ((b^3)/6)x^3+x^2+x+1
So Max[f(1)]=b/6+1+1+1<5 > b<=12
I'm not a math major student. If you think my solution is not convinced, you can prove it yourself..=_=iiii
Re: URGENT GRE 9367 Q42 AND Q23
define g(x)=f(x)x which is continuous on [0,1]
g(0)=f(0)0=1>0
g(1)=f(1)1=01=1<0
By IVT,
There is a x in (0,1) such that g(x)=f(x)x=0
=>
f(x)=x
Wishing u all the best for your GRE.
g(0)=f(0)0=1>0
g(1)=f(1)1=01=1<0
By IVT,
There is a x in (0,1) such that g(x)=f(x)x=0
=>
f(x)=x
Wishing u all the best for your GRE.

 Posts: 5
 Joined: Tue Nov 01, 2011 5:44 am
Re: URGENT GRE 9367 Q42 AND Q23
Thanks a lot... that cleared my concepts regarding IVT.... just one more favour....yoyobarn wrote:define g(x)=f(x)x which is continuous on [0,1]
g(0)=f(0)0=1>0
g(1)=f(1)1=01=1<0
By IVT,
There is a x in (0,1) such that g(x)=f(x)x=0
=>
f(x)=x
Wishing u all the best for your GRE.
can you solve Q42 without use of mathematica....... just for curiosity...

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: URGENT GRE 9367 Q42 AND Q23
For 42, the worst case scenario is that your 3rd derivative is constantly equal to B. This is the case for the poly (B/6)x^3+x^2+x+1. (which satisfies the other conditions). Plug in one, so you get 3+B/6 which needs to be less than 5, thus B is less than 12

 Posts: 5
 Joined: Tue Nov 01, 2011 5:44 am
Re: URGENT GRE 9367 Q42 AND Q23
Thanks again.... your help means a lot...blitzer6266 wrote:For 42, the worst case scenario is that your 3rd derivative is constantly equal to B. This is the case for the poly (B/6)x^3+x^2+x+1. (which satisfies the other conditions). Plug in one, so you get 3+B/6 which needs to be less than 5, thus B is less than 12
Re: URGENT GRE 9367 Q42 AND Q23
Well, just wiki it http://en.wikipedia.org/wiki/Intermediate_value_theorem...lifeisgood88 wrote:Thanks a lot... but can you explain me the intermediate value theorem part.... i am bit confused about it... as it states that if f(x) is continuous on [a,b] and k is any number between f(a) and f(b), then there exists at least one no. c such that a<c<b and f(c)=k.mhyyh wrote:23.lifeisgood88 wrote:42. What is the greatest value of b for which any real valued function f that satisfies the following properties must also satisfy f(1)<5 ?
1> f is infinitely differentiable on the real numbers
2> f(0)=1, f'(0)=1, and f''(0)=2; and
3> f'''(x)< b for all x in [0,1]
a)1
b)2
c)6
d)12
e)24
23.Let f be a real valued function continuous on the close interval [0,1] and differentiable on the open interval (0,1) with f(0)=1 and f(1)=0. Which of the following must be true
1. there exists x in (0,1) such that f(x)=x
2. there exists x in (0,1) such that f'(x)=1
3. f(x)>0 for all x in [0,1)
a) 1 only
b) 2 only
c) 1 and 2 only
d) 2 and 3 only
e) 1.2 and 3
Please post as soon as possible. I am in a big mess regarding this questions. Please Help.
according to intermediate value theorem, 1 is obvious.
according to mean value theorem, 2 is obvious.
and the last one is still obvious.....show u a example:y=2x^23x+1
42.(I hope you know the language of Wolfram's softwareMathematica...)
Integrate[f'''(x),{x,0,1}]=f''(1)f''(0), and so do f''(x) and f'(x)
5>Max[f(1)]=Max[Integrate[f'(x),{x,0,1}]+f(0)=Max[Integrate[f''(x),{x,0,1}]+f'(0)+f(0)=Max[Integrate[f'''(x),{x,0,1}]+f''(0)+f'(0)+f(0)
Apparently, Max[f'''(x)]=b > Max[f''(x)]=bx+2 > Max[f'(x)]=(b/2)*x^2+2x+1 > Max[f(x)]= ((b^3)/6)x^3+x^2+x+1
So Max[f(1)]=b/6+1+1+1<5 > b<=12
I'm not a math major student. If you think my solution is not convinced, you can prove it yourself..=_=iiii
f() is a continuous function on [a,b].
for every min{f(a),f(b)}<=y_0<=max{f(a),f(b)}
there exist at least one x from [a,b]
s.t. f(x_0)=y_0..

 Posts: 2
 Joined: Fri Mar 21, 2014 2:05 am
Re: URGENT GRE 9367 Q42 AND Q23
For (I), the intermediate value theorem only works in a closed interval, no?
If I take f(x)=x^2+1, there is no x in an open interval (0, 1) where f(x)=x.
Can someone please clarify?
If I take f(x)=x^2+1, there is no x in an open interval (0, 1) where f(x)=x.
Can someone please clarify?
Last edited by orangebaby on Fri Mar 21, 2014 3:33 am, edited 1 time in total.
Re: URGENT GRE 9367 Q42 AND Q23
The answer is right above you, can you not even be bothered by at least looking at the post?

 Posts: 18
 Joined: Tue Oct 29, 2013 10:29 am
Re: URGENT GRE 9367 Q42 AND Q23
Sure there is. Find the intersection of the curves y = x^2 + 1 and y = x.orangebaby wrote:For (I), the intermediate value theorem only works in a closed interval, no?
If I take f(x)=x^2+1, there is no x in an open interval (0, 1) where f(x)=x.
Can someone please clarify?
x is approximately .62.
Re: URGENT GRE 9367 Q42 AND Q23
I know this question post is old, but I think this reply will be helpful to anyone thinking about this problem.
Question 42 is essentially asking you about Taylor's theorem.
(I had originally read f''(x) < b in the problem statement and later learned it was a triple derivative, whence I realized it was a simple question on Taylor's theorem.)
Question 42 is essentially asking you about Taylor's theorem.
(I had originally read f''(x) < b in the problem statement and later learned it was a triple derivative, whence I realized it was a simple question on Taylor's theorem.)