Prove $ (b-1)e^{-b} < (a-1)e^{-a} $ for every $ 2 < a < b $
$begingroup$
Prove: for every $ 2 < a < b $ : $ (b-1)e^{-b} < (a-1)e^{-a} $
My first try was to try and build a function and show it is going down:
$ f(x) = (b-1)e^{-b} - (a-1)e^{-a} $
but I don't know how to handle such function. Is there any easier way?
calculus
asked Jan 6 at 16:30
bm1125bm1125
64916
$endgroup$
add a comment |
$begingroup$
Prove: for every $ 2 < a < b $ : $ (b-1)e^{-b} < (a-1)e^{-a} $
My first try was to try and build a function and show it is going down:
$ f(x) = (b-1)e^{-b} - (a-1)e^{-a} $
but I don't know how to handle such function. Is there any easier way?
calculus
asked Jan 6 at 16:30
bm1125bm1125
64916
$endgroup$
$begingroup$
Where is the variable $x$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 16:33
add a comment |
$begingroup$
Prove: for every $ 2 < a < b $ : $ (b-1)e^{-b} < (a-1)e^{-a} $
My first try was to try and build a function and show it is going down:
$ f(x) = (b-1)e^{-b} - (a-1)e^{-a} $
but I don't know how to handle such function. Is there any easier way?
calculus
asked Jan 6 at 16:30
bm1125bm1125
64916
$endgroup$
Prove: for every $ 2 < a < b $ : $ (b-1)e^{-b} < (a-1)e^{-a} $
My first try was to try and build a function and show it is going down:
$ f(x) = (b-1)e^{-b} - (a-1)e^{-a} $
but I don't know how to handle such function. Is there any easier way?
calculus
calculus
asked Jan 6 at 16:30
bm1125bm1125
64916
asked Jan 6 at 16:30
bm1125bm1125
64916
asked Jan 6 at 16:30
bm1125bm1125
64916
asked Jan 6 at 16:30
bm1125bm1125
64916
asked Jan 6 at 16:30
bm1125bm1125
64916
64916
$begingroup$
Where is the variable $x$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 16:33
add a comment |
$begingroup$
Where is the variable $x$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 16:33
$begingroup$
Where is the variable $x$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 16:33
$begingroup$
Where is the variable $x$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 16:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Take
$$f(x) = (x-1)e^{-x}$$
then
$$f'(x) = e^{-x}-e^{-x}(x-1)=e^{-x}(2-x)<0$$
when $x>2$ and so $f$ is decreasing when $x>2.$ Thus for $2<a<b$ we have that
$$f(a)>f(b)implies (b-1)e^{-b} < (a-1)e^{-a}.$$
answered Jan 6 at 16:37
Hello_WorldHello_World
4,13121831
$endgroup$
add a comment |
$begingroup$
Hint/Guide
Consider the function
$$
f(x)=frac{x-1}{e^x}.
$$
By computing its derivative, show that $f$ is decreasing for $x>2$ from which the claim will follow.
answered Jan 6 at 16:34
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take
$$f(x) = (x-1)e^{-x}$$
then
$$f'(x) = e^{-x}-e^{-x}(x-1)=e^{-x}(2-x)<0$$
when $x>2$ and so $f$ is decreasing when $x>2.$ Thus for $2<a<b$ we have that
$$f(a)>f(b)implies (b-1)e^{-b} < (a-1)e^{-a}.$$
answered Jan 6 at 16:37
Hello_WorldHello_World
4,13121831
$endgroup$
add a comment |
$begingroup$
Take
$$f(x) = (x-1)e^{-x}$$
then
$$f'(x) = e^{-x}-e^{-x}(x-1)=e^{-x}(2-x)<0$$
when $x>2$ and so $f$ is decreasing when $x>2.$ Thus for $2<a<b$ we have that
$$f(a)>f(b)implies (b-1)e^{-b} < (a-1)e^{-a}.$$
answered Jan 6 at 16:37
Hello_WorldHello_World
4,13121831
$endgroup$
add a comment |
$begingroup$
Take
$$f(x) = (x-1)e^{-x}$$
then
$$f'(x) = e^{-x}-e^{-x}(x-1)=e^{-x}(2-x)<0$$
when $x>2$ and so $f$ is decreasing when $x>2.$ Thus for $2<a<b$ we have that
$$f(a)>f(b)implies (b-1)e^{-b} < (a-1)e^{-a}.$$
answered Jan 6 at 16:37
Hello_WorldHello_World
4,13121831
$endgroup$
Take
$$f(x) = (x-1)e^{-x}$$
then
$$f'(x) = e^{-x}-e^{-x}(x-1)=e^{-x}(2-x)<0$$
when $x>2$ and so $f$ is decreasing when $x>2.$ Thus for $2<a<b$ we have that
$$f(a)>f(b)implies (b-1)e^{-b} < (a-1)e^{-a}.$$
answered Jan 6 at 16:37
Hello_WorldHello_World
4,13121831
answered Jan 6 at 16:37
Hello_WorldHello_World
4,13121831
answered Jan 6 at 16:37
Hello_WorldHello_World
4,13121831
answered Jan 6 at 16:37
Hello_WorldHello_World
4,13121831
4,13121831
add a comment |
add a comment |
$begingroup$
Hint/Guide
Consider the function
$$
f(x)=frac{x-1}{e^x}.
$$
By computing its derivative, show that $f$ is decreasing for $x>2$ from which the claim will follow.
answered Jan 6 at 16:34
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
add a comment |
$begingroup$
Hint/Guide
Consider the function
$$
f(x)=frac{x-1}{e^x}.
$$
By computing its derivative, show that $f$ is decreasing for $x>2$ from which the claim will follow.
answered Jan 6 at 16:34
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
add a comment |
$begingroup$
Hint/Guide
Consider the function
$$
f(x)=frac{x-1}{e^x}.
$$
By computing its derivative, show that $f$ is decreasing for $x>2$ from which the claim will follow.
answered Jan 6 at 16:34
Foobaz JohnFoobaz John
22.1k41352
$endgroup$
Hint/Guide
Consider the function
$$
f(x)=frac{x-1}{e^x}.
$$
By computing its derivative, show that $f$ is decreasing for $x>2$ from which the claim will follow.
answered Jan 6 at 16:34
Foobaz JohnFoobaz John
22.1k41352
answered Jan 6 at 16:34
Foobaz JohnFoobaz John
22.1k41352
answered Jan 6 at 16:34
Foobaz JohnFoobaz John
22.1k41352
answered Jan 6 at 16:34
Foobaz JohnFoobaz John
22.1k41352
22.1k41352
add a comment |
add a comment |
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$begingroup$
Where is the variable $x$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 16:33