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Calculus: Mean Value Theorems
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Suppose $f(x)$ is twice differentiable for all real $x$. Further suppose $|f(x)|leqslant$ 1 and $|f''(x)| leqslant$ 1 for all real $x$. Then $|f'(x)|$ is:
Options:
a) $leqslant$ 1$quad$
b) $gt$ 2 $quad$
c) $leqslant frac 32$ $quad$
d) $gt frac 32$
My thoughts:
I first thought it could be $sin(x)$ or $cos(x)$ as it satisfied all the given conditions. I looked up the answer, and it was c. Since I had information about $f$ and $f''$ I tried applying Lagrange's Mean Value theorem on $f(x)$ and $f'(x)$ assuming arbitrary parameters to get two inequalities. $$fracf(b)-f(a)b-a = f'(c)$$ for some arbitrary $a,b$ and $c$, and $c ; epsilon ; (a,b)$. Similarly I applied the same theorem for $f'(x)$. Using the given conditions I could get inequalities using the given conditions on the values of $f(x)$ and $f''(x)$. I could not proceed further.
calculus derivatives
asked Jul 30 at 15:54
Shreyas K
92
add a comment |Â
up vote
-1
down vote
favorite
Suppose $f(x)$ is twice differentiable for all real $x$. Further suppose $|f(x)|leqslant$ 1 and $|f''(x)| leqslant$ 1 for all real $x$. Then $|f'(x)|$ is:
Options:
a) $leqslant$ 1$quad$
b) $gt$ 2 $quad$
c) $leqslant frac 32$ $quad$
d) $gt frac 32$
My thoughts:
I first thought it could be $sin(x)$ or $cos(x)$ as it satisfied all the given conditions. I looked up the answer, and it was c. Since I had information about $f$ and $f''$ I tried applying Lagrange's Mean Value theorem on $f(x)$ and $f'(x)$ assuming arbitrary parameters to get two inequalities. $$fracf(b)-f(a)b-a = f'(c)$$ for some arbitrary $a,b$ and $c$, and $c ; epsilon ; (a,b)$. Similarly I applied the same theorem for $f'(x)$. Using the given conditions I could get inequalities using the given conditions on the values of $f(x)$ and $f''(x)$. I could not proceed further.
calculus derivatives
asked Jul 30 at 15:54
Shreyas K
92
I tried applying Lagrange's mean value theorem on f'(x), but it didn't lead me anywhere useful.
– Shreyas K
Jul 30 at 15:58
Taylor theorem.
– xbh
Jul 30 at 15:58
1
This is not a proper method but you can assume the function to be $sin(x)$. It satisfies the given conditions and so the answer should be (a). What's the answer?
– Alphanerd
Jul 30 at 16:59
The answer is C
– Shreyas K
Jul 31 at 11:31
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Suppose $f(x)$ is twice differentiable for all real $x$. Further suppose $|f(x)|leqslant$ 1 and $|f''(x)| leqslant$ 1 for all real $x$. Then $|f'(x)|$ is:
Options:
a) $leqslant$ 1$quad$
b) $gt$ 2 $quad$
c) $leqslant frac 32$ $quad$
d) $gt frac 32$
My thoughts:
I first thought it could be $sin(x)$ or $cos(x)$ as it satisfied all the given conditions. I looked up the answer, and it was c. Since I had information about $f$ and $f''$ I tried applying Lagrange's Mean Value theorem on $f(x)$ and $f'(x)$ assuming arbitrary parameters to get two inequalities. $$fracf(b)-f(a)b-a = f'(c)$$ for some arbitrary $a,b$ and $c$, and $c ; epsilon ; (a,b)$. Similarly I applied the same theorem for $f'(x)$. Using the given conditions I could get inequalities using the given conditions on the values of $f(x)$ and $f''(x)$. I could not proceed further.
calculus derivatives
asked Jul 30 at 15:54
Shreyas K
92
Suppose $f(x)$ is twice differentiable for all real $x$. Further suppose $|f(x)|leqslant$ 1 and $|f''(x)| leqslant$ 1 for all real $x$. Then $|f'(x)|$ is:
Options:
a) $leqslant$ 1$quad$
b) $gt$ 2 $quad$
c) $leqslant frac 32$ $quad$
d) $gt frac 32$
My thoughts:
I first thought it could be $sin(x)$ or $cos(x)$ as it satisfied all the given conditions. I looked up the answer, and it was c. Since I had information about $f$ and $f''$ I tried applying Lagrange's Mean Value theorem on $f(x)$ and $f'(x)$ assuming arbitrary parameters to get two inequalities. $$fracf(b)-f(a)b-a = f'(c)$$ for some arbitrary $a,b$ and $c$, and $c ; epsilon ; (a,b)$. Similarly I applied the same theorem for $f'(x)$. Using the given conditions I could get inequalities using the given conditions on the values of $f(x)$ and $f''(x)$. I could not proceed further.
calculus derivatives
asked Jul 30 at 15:54
Shreyas K
92
edited Jul 31 at 11:31
asked Jul 30 at 15:54
Shreyas K
92
asked Jul 30 at 15:54
Shreyas K
92
asked Jul 30 at 15:54
Shreyas K
92
92
I tried applying Lagrange's mean value theorem on f'(x), but it didn't lead me anywhere useful.
– Shreyas K
Jul 30 at 15:58
Taylor theorem.
– xbh
Jul 30 at 15:58
1
This is not a proper method but you can assume the function to be $sin(x)$. It satisfies the given conditions and so the answer should be (a). What's the answer?
– Alphanerd
Jul 30 at 16:59
The answer is C
– Shreyas K
Jul 31 at 11:31
add a comment |Â
I tried applying Lagrange's mean value theorem on f'(x), but it didn't lead me anywhere useful.
– Shreyas K
Jul 30 at 15:58
Taylor theorem.
– xbh
Jul 30 at 15:58
1
This is not a proper method but you can assume the function to be $sin(x)$. It satisfies the given conditions and so the answer should be (a). What's the answer?
– Alphanerd
Jul 30 at 16:59
The answer is C
– Shreyas K
Jul 31 at 11:31
I tried applying Lagrange's mean value theorem on f'(x), but it didn't lead me anywhere useful.
– Shreyas K
Jul 30 at 15:58
I tried applying Lagrange's mean value theorem on f'(x), but it didn't lead me anywhere useful.
– Shreyas K
Jul 30 at 15:58
Taylor theorem.
– xbh
Jul 30 at 15:58
Taylor theorem.
– xbh
Jul 30 at 15:58
1
1
This is not a proper method but you can assume the function to be $sin(x)$. It satisfies the given conditions and so the answer should be (a). What's the answer?
– Alphanerd
Jul 30 at 16:59
This is not a proper method but you can assume the function to be $sin(x)$. It satisfies the given conditions and so the answer should be (a). What's the answer?
– Alphanerd
Jul 30 at 16:59
The answer is C
– Shreyas K
Jul 31 at 11:31
The answer is C
– Shreyas K
Jul 31 at 11:31
add a comment |Â
1 Answer
1
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1
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As I commented, the key is Taylor's theorem, not the MVT.
Solution.$blacktriangleleft$. For an arbitrary $h >0$, consider Taylor's formula with Lagrange's remainder:
beginalign*
f(x + h) &= f(x) + h f'(x) + frac h^22 f''(c_1)quad [c_1 in (x, x+h)],\
f(x - h) &= f(x) - hf'(x) + frac h^22 f''(c_2) quad [c_2 in (x-h, x)].
endalign*
Subtracting them yields
$$
f(x+h) - f(x-h) = 2h f'(x) + frac h^22 (f''(c_1) - f''(c_2)).
$$
Now estimate $|f'(x)|$:
beginalign*
|f'(x)| &= left|frac f(x+h) - f(x-h) 2h + frac h 4 (f''(c_1) - f''(c_2))right|\
&leqslant frac f(x+h) 2h + frac h 4 (|f''(c_1)| + |f''(c_2)|) \
&leqslant frac 22h + frac h 4 cdot 2\
&= frac 1h + frac h2.
endalign*
Therefore,
$$
sup_xin mathbb R|f'(x)| leqslant frac 1h + frac h2 quad [forall h > 0].
$$
By AM-GM inequality, the RHS has a minimum $2 sqrt(1/h) cdot (h/2) = sqrt2$, thus
$$
|f'(x)| leqslant sqrt 2 doteq 1.414 fbox$leqslant frac 32$,
$$
hence the choice $(mathrm C)$. $blacktriangleright$
answered Aug 1 at 3:53
xbh
1,0257
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
As I commented, the key is Taylor's theorem, not the MVT.
Solution.$blacktriangleleft$. For an arbitrary $h >0$, consider Taylor's formula with Lagrange's remainder:
beginalign*
f(x + h) &= f(x) + h f'(x) + frac h^22 f''(c_1)quad [c_1 in (x, x+h)],\
f(x - h) &= f(x) - hf'(x) + frac h^22 f''(c_2) quad [c_2 in (x-h, x)].
endalign*
Subtracting them yields
$$
f(x+h) - f(x-h) = 2h f'(x) + frac h^22 (f''(c_1) - f''(c_2)).
$$
Now estimate $|f'(x)|$:
beginalign*
|f'(x)| &= left|frac f(x+h) - f(x-h) 2h + frac h 4 (f''(c_1) - f''(c_2))right|\
&leqslant frac f(x+h) 2h + frac h 4 (|f''(c_1)| + |f''(c_2)|) \
&leqslant frac 22h + frac h 4 cdot 2\
&= frac 1h + frac h2.
endalign*
Therefore,
$$
sup_xin mathbb R|f'(x)| leqslant frac 1h + frac h2 quad [forall h > 0].
$$
By AM-GM inequality, the RHS has a minimum $2 sqrt(1/h) cdot (h/2) = sqrt2$, thus
$$
|f'(x)| leqslant sqrt 2 doteq 1.414 fbox$leqslant frac 32$,
$$
hence the choice $(mathrm C)$. $blacktriangleright$
answered Aug 1 at 3:53
xbh
1,0257
add a comment |Â
up vote
1
down vote
As I commented, the key is Taylor's theorem, not the MVT.
Solution.$blacktriangleleft$. For an arbitrary $h >0$, consider Taylor's formula with Lagrange's remainder:
beginalign*
f(x + h) &= f(x) + h f'(x) + frac h^22 f''(c_1)quad [c_1 in (x, x+h)],\
f(x - h) &= f(x) - hf'(x) + frac h^22 f''(c_2) quad [c_2 in (x-h, x)].
endalign*
Subtracting them yields
$$
f(x+h) - f(x-h) = 2h f'(x) + frac h^22 (f''(c_1) - f''(c_2)).
$$
Now estimate $|f'(x)|$:
beginalign*
|f'(x)| &= left|frac f(x+h) - f(x-h) 2h + frac h 4 (f''(c_1) - f''(c_2))right|\
&leqslant frac f(x+h) 2h + frac h 4 (|f''(c_1)| + |f''(c_2)|) \
&leqslant frac 22h + frac h 4 cdot 2\
&= frac 1h + frac h2.
endalign*
Therefore,
$$
sup_xin mathbb R|f'(x)| leqslant frac 1h + frac h2 quad [forall h > 0].
$$
By AM-GM inequality, the RHS has a minimum $2 sqrt(1/h) cdot (h/2) = sqrt2$, thus
$$
|f'(x)| leqslant sqrt 2 doteq 1.414 fbox$leqslant frac 32$,
$$
hence the choice $(mathrm C)$. $blacktriangleright$
answered Aug 1 at 3:53
xbh
1,0257
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As I commented, the key is Taylor's theorem, not the MVT.
Solution.$blacktriangleleft$. For an arbitrary $h >0$, consider Taylor's formula with Lagrange's remainder:
beginalign*
f(x + h) &= f(x) + h f'(x) + frac h^22 f''(c_1)quad [c_1 in (x, x+h)],\
f(x - h) &= f(x) - hf'(x) + frac h^22 f''(c_2) quad [c_2 in (x-h, x)].
endalign*
Subtracting them yields
$$
f(x+h) - f(x-h) = 2h f'(x) + frac h^22 (f''(c_1) - f''(c_2)).
$$
Now estimate $|f'(x)|$:
beginalign*
|f'(x)| &= left|frac f(x+h) - f(x-h) 2h + frac h 4 (f''(c_1) - f''(c_2))right|\
&leqslant frac f(x+h) 2h + frac h 4 (|f''(c_1)| + |f''(c_2)|) \
&leqslant frac 22h + frac h 4 cdot 2\
&= frac 1h + frac h2.
endalign*
Therefore,
$$
sup_xin mathbb R|f'(x)| leqslant frac 1h + frac h2 quad [forall h > 0].
$$
By AM-GM inequality, the RHS has a minimum $2 sqrt(1/h) cdot (h/2) = sqrt2$, thus
$$
|f'(x)| leqslant sqrt 2 doteq 1.414 fbox$leqslant frac 32$,
$$
hence the choice $(mathrm C)$. $blacktriangleright$
answered Aug 1 at 3:53
xbh
1,0257
As I commented, the key is Taylor's theorem, not the MVT.
Solution.$blacktriangleleft$. For an arbitrary $h >0$, consider Taylor's formula with Lagrange's remainder:
beginalign*
f(x + h) &= f(x) + h f'(x) + frac h^22 f''(c_1)quad [c_1 in (x, x+h)],\
f(x - h) &= f(x) - hf'(x) + frac h^22 f''(c_2) quad [c_2 in (x-h, x)].
endalign*
Subtracting them yields
$$
f(x+h) - f(x-h) = 2h f'(x) + frac h^22 (f''(c_1) - f''(c_2)).
$$
Now estimate $|f'(x)|$:
beginalign*
|f'(x)| &= left|frac f(x+h) - f(x-h) 2h + frac h 4 (f''(c_1) - f''(c_2))right|\
&leqslant frac f(x+h) 2h + frac h 4 (|f''(c_1)| + |f''(c_2)|) \
&leqslant frac 22h + frac h 4 cdot 2\
&= frac 1h + frac h2.
endalign*
Therefore,
$$
sup_xin mathbb R|f'(x)| leqslant frac 1h + frac h2 quad [forall h > 0].
$$
By AM-GM inequality, the RHS has a minimum $2 sqrt(1/h) cdot (h/2) = sqrt2$, thus
$$
|f'(x)| leqslant sqrt 2 doteq 1.414 fbox$leqslant frac 32$,
$$
hence the choice $(mathrm C)$. $blacktriangleright$
answered Aug 1 at 3:53
xbh
1,0257
answered Aug 1 at 3:53
xbh
1,0257
answered Aug 1 at 3:53
xbh
1,0257
answered Aug 1 at 3:53
xbh
1,0257
1,0257
add a comment |Â
add a comment |Â
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I tried applying Lagrange's mean value theorem on f'(x), but it didn't lead me anywhere useful.
– Shreyas K
Jul 30 at 15:58
Taylor theorem.
– xbh
Jul 30 at 15:58
1
This is not a proper method but you can assume the function to be $sin(x)$. It satisfies the given conditions and so the answer should be (a). What's the answer?
– Alphanerd
Jul 30 at 16:59
The answer is C
– Shreyas K
Jul 31 at 11:31