$ex leq e^x$ inequality using derivatives
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So I was going through a sum for
Prove $ex leq e^x$ , $forall x in mathbb{R} $
I took $g(x) = e^x - ex$
Then $g'(x)= e^x - e$
I understood the case when $x>1$ function is strictly increasing i.e $g'(x) >0$ then $e^x>ex$ but what about when $x leq 1$
calculus
asked Nov 24 at 16:29
Sumukh Sai
206
add a comment |
up vote
0
down vote
favorite
So I was going through a sum for
Prove $ex leq e^x$ , $forall x in mathbb{R} $
I took $g(x) = e^x - ex$
Then $g'(x)= e^x - e$
I understood the case when $x>1$ function is strictly increasing i.e $g'(x) >0$ then $e^x>ex$ but what about when $x leq 1$
calculus
asked Nov 24 at 16:29
Sumukh Sai
206
3
Then the function is decreasing. That's all you need. Think about it for a moment.
– saulspatz
Nov 24 at 16:32
Just re-wording what @saulspatz says: what does it mean for a function to have only one minima or maxima? Write it as inequality
– Andrei
Nov 24 at 16:35
I'm sorry I'm still getting a bit confused
– Sumukh Sai
Nov 24 at 16:45
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So I was going through a sum for
Prove $ex leq e^x$ , $forall x in mathbb{R} $
I took $g(x) = e^x - ex$
Then $g'(x)= e^x - e$
I understood the case when $x>1$ function is strictly increasing i.e $g'(x) >0$ then $e^x>ex$ but what about when $x leq 1$
calculus
asked Nov 24 at 16:29
Sumukh Sai
206
So I was going through a sum for
Prove $ex leq e^x$ , $forall x in mathbb{R} $
I took $g(x) = e^x - ex$
Then $g'(x)= e^x - e$
I understood the case when $x>1$ function is strictly increasing i.e $g'(x) >0$ then $e^x>ex$ but what about when $x leq 1$
calculus
calculus
asked Nov 24 at 16:29
Sumukh Sai
206
asked Nov 24 at 16:29
Sumukh Sai
206
asked Nov 24 at 16:29
Sumukh Sai
206
asked Nov 24 at 16:29
Sumukh Sai
206
asked Nov 24 at 16:29
Sumukh Sai
206
206
3
Then the function is decreasing. That's all you need. Think about it for a moment.
– saulspatz
Nov 24 at 16:32
Just re-wording what @saulspatz says: what does it mean for a function to have only one minima or maxima? Write it as inequality
– Andrei
Nov 24 at 16:35
I'm sorry I'm still getting a bit confused
– Sumukh Sai
Nov 24 at 16:45
add a comment |
3
Then the function is decreasing. That's all you need. Think about it for a moment.
– saulspatz
Nov 24 at 16:32
Just re-wording what @saulspatz says: what does it mean for a function to have only one minima or maxima? Write it as inequality
– Andrei
Nov 24 at 16:35
I'm sorry I'm still getting a bit confused
– Sumukh Sai
Nov 24 at 16:45
3
3
Then the function is decreasing. That's all you need. Think about it for a moment.
– saulspatz
Nov 24 at 16:32
Then the function is decreasing. That's all you need. Think about it for a moment.
– saulspatz
Nov 24 at 16:32
Just re-wording what @saulspatz says: what does it mean for a function to have only one minima or maxima? Write it as inequality
– Andrei
Nov 24 at 16:35
Just re-wording what @saulspatz says: what does it mean for a function to have only one minima or maxima? Write it as inequality
– Andrei
Nov 24 at 16:35
I'm sorry I'm still getting a bit confused
– Sumukh Sai
Nov 24 at 16:45
I'm sorry I'm still getting a bit confused
– Sumukh Sai
Nov 24 at 16:45
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
it is easy to see $x=1$ is the absolute minimum, that is, $ f(1)leq f(x)$ for all $xin R$ so that $0leq ex-e^x$.Thus $exleq e^x$ for all $xin R.$
answered Nov 24 at 17:07
gb2017
944
add a comment |
up vote
0
down vote
We have that
$$g(x)=e^x-ex implies g'(x)=e^x-e=0quad x=1$$
moreover $g''(x)=e^x > 0$, therefore $x=1$ is a point of minimum for $g(x)$.
answered Nov 24 at 16:57
gimusi
90.7k74495
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
it is easy to see $x=1$ is the absolute minimum, that is, $ f(1)leq f(x)$ for all $xin R$ so that $0leq ex-e^x$.Thus $exleq e^x$ for all $xin R.$
answered Nov 24 at 17:07
gb2017
944
add a comment |
up vote
1
down vote
accepted
it is easy to see $x=1$ is the absolute minimum, that is, $ f(1)leq f(x)$ for all $xin R$ so that $0leq ex-e^x$.Thus $exleq e^x$ for all $xin R.$
answered Nov 24 at 17:07
gb2017
944
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
it is easy to see $x=1$ is the absolute minimum, that is, $ f(1)leq f(x)$ for all $xin R$ so that $0leq ex-e^x$.Thus $exleq e^x$ for all $xin R.$
answered Nov 24 at 17:07
gb2017
944
it is easy to see $x=1$ is the absolute minimum, that is, $ f(1)leq f(x)$ for all $xin R$ so that $0leq ex-e^x$.Thus $exleq e^x$ for all $xin R.$
answered Nov 24 at 17:07
gb2017
944
answered Nov 24 at 17:07
gb2017
944
answered Nov 24 at 17:07
gb2017
944
answered Nov 24 at 17:07
gb2017
944
944
add a comment |
add a comment |
up vote
0
down vote
We have that
$$g(x)=e^x-ex implies g'(x)=e^x-e=0quad x=1$$
moreover $g''(x)=e^x > 0$, therefore $x=1$ is a point of minimum for $g(x)$.
answered Nov 24 at 16:57
gimusi
90.7k74495
add a comment |
up vote
0
down vote
We have that
$$g(x)=e^x-ex implies g'(x)=e^x-e=0quad x=1$$
moreover $g''(x)=e^x > 0$, therefore $x=1$ is a point of minimum for $g(x)$.
answered Nov 24 at 16:57
gimusi
90.7k74495
add a comment |
up vote
0
down vote
up vote
0
down vote
We have that
$$g(x)=e^x-ex implies g'(x)=e^x-e=0quad x=1$$
moreover $g''(x)=e^x > 0$, therefore $x=1$ is a point of minimum for $g(x)$.
answered Nov 24 at 16:57
gimusi
90.7k74495
We have that
$$g(x)=e^x-ex implies g'(x)=e^x-e=0quad x=1$$
moreover $g''(x)=e^x > 0$, therefore $x=1$ is a point of minimum for $g(x)$.
answered Nov 24 at 16:57
gimusi
90.7k74495
edited Nov 24 at 17:25
answered Nov 24 at 16:57
gimusi
90.7k74495
answered Nov 24 at 16:57
gimusi
90.7k74495
answered Nov 24 at 16:57
gimusi
90.7k74495
90.7k74495
add a comment |
add a comment |
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3
Then the function is decreasing. That's all you need. Think about it for a moment.
– saulspatz
Nov 24 at 16:32
Just re-wording what @saulspatz says: what does it mean for a function to have only one minima or maxima? Write it as inequality
– Andrei
Nov 24 at 16:35
I'm sorry I'm still getting a bit confused
– Sumukh Sai
Nov 24 at 16:45