how to prove that exp(x)>x+1 [duplicate]
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This question already has an answer here:
Simplest or nicest proof that $1+x le e^x$
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- Hey i want the easiest method to prove exp(x)>x+1
The only method i use is to consider a new function F, that realizes
F(x)=exp(x)-x-1 then calculate the derivative then use its monotony to prove that F(x)<0
I'm only a high school student,if you could use function study it would be easier for me to understand
So do you have any better method? this one takes me some time to write down i would love a easier method
calculus inequality exponential-function
asked Dec 31 '18 at 13:07
Bénz AnasBénz Anas
1
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marked as duplicate by José Carlos Santos calculus
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Dec 31 '18 at 13:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
Simplest or nicest proof that $1+x le e^x$
24 answers
- Hey i want the easiest method to prove exp(x)>x+1
The only method i use is to consider a new function F, that realizes
F(x)=exp(x)-x-1 then calculate the derivative then use its monotony to prove that F(x)<0
I'm only a high school student,if you could use function study it would be easier for me to understand
So do you have any better method? this one takes me some time to write down i would love a easier method
calculus inequality exponential-function
asked Dec 31 '18 at 13:07
Bénz AnasBénz Anas
1
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marked as duplicate by José Carlos Santos calculus
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Dec 31 '18 at 13:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Simplest or nicest proof that $1+x le e^x$
24 answers
- Hey i want the easiest method to prove exp(x)>x+1
The only method i use is to consider a new function F, that realizes
F(x)=exp(x)-x-1 then calculate the derivative then use its monotony to prove that F(x)<0
I'm only a high school student,if you could use function study it would be easier for me to understand
So do you have any better method? this one takes me some time to write down i would love a easier method
calculus inequality exponential-function
asked Dec 31 '18 at 13:07
Bénz AnasBénz Anas
1
$endgroup$
This question already has an answer here:
Simplest or nicest proof that $1+x le e^x$
24 answers
- Hey i want the easiest method to prove exp(x)>x+1
The only method i use is to consider a new function F, that realizes
F(x)=exp(x)-x-1 then calculate the derivative then use its monotony to prove that F(x)<0
I'm only a high school student,if you could use function study it would be easier for me to understand
So do you have any better method? this one takes me some time to write down i would love a easier method
This question already has an answer here:
Simplest or nicest proof that $1+x le e^x$
24 answers
calculus inequality exponential-function
calculus inequality exponential-function
asked Dec 31 '18 at 13:07
Bénz AnasBénz Anas
1
asked Dec 31 '18 at 13:07
Bénz AnasBénz Anas
1
asked Dec 31 '18 at 13:07
Bénz AnasBénz Anas
1
asked Dec 31 '18 at 13:07
Bénz AnasBénz Anas
1
asked Dec 31 '18 at 13:07
Bénz AnasBénz Anas
1
1
marked as duplicate by José Carlos Santos calculus
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Dec 31 '18 at 13:10
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Dec 31 '18 at 13:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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2 Answers
2
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oldest
votes
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$$exp(x) = 1 + int_0^{x} exp(t) mathrm dt > 1 + int_0^x 1 mathrm dt = 1 + x$$
answered Dec 31 '18 at 13:09
Kenny LauKenny Lau
19.9k2159
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That's what i was looking for! Beautiful method,thanks for taking the time to share it with me i'm grateful.
$endgroup$
– Bénz Anas
Jan 1 at 10:51
add a comment |
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Defining $$f(x)=e^x-x-1$$ then $$f'(x)=e^x-1$$ and $$f''(x)=e^x>0$$ Can you proceed?
answered Dec 31 '18 at 13:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42865
$endgroup$
1
$begingroup$
Sometimes I wonder how much of the questions you read. OP wrote: "The only method i use is to consider a new function F, that realizes F(x)=exp(x)-x-1 then calculate the derivative then use its monotony ... So do you have any better method?"
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– Martin R
Dec 31 '18 at 18:50
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
$$exp(x) = 1 + int_0^{x} exp(t) mathrm dt > 1 + int_0^x 1 mathrm dt = 1 + x$$
answered Dec 31 '18 at 13:09
Kenny LauKenny Lau
19.9k2159
$endgroup$
$begingroup$
That's what i was looking for! Beautiful method,thanks for taking the time to share it with me i'm grateful.
$endgroup$
– Bénz Anas
Jan 1 at 10:51
add a comment |
$begingroup$
$$exp(x) = 1 + int_0^{x} exp(t) mathrm dt > 1 + int_0^x 1 mathrm dt = 1 + x$$
answered Dec 31 '18 at 13:09
Kenny LauKenny Lau
19.9k2159
$endgroup$
$begingroup$
That's what i was looking for! Beautiful method,thanks for taking the time to share it with me i'm grateful.
$endgroup$
– Bénz Anas
Jan 1 at 10:51
add a comment |
$begingroup$
$$exp(x) = 1 + int_0^{x} exp(t) mathrm dt > 1 + int_0^x 1 mathrm dt = 1 + x$$
answered Dec 31 '18 at 13:09
Kenny LauKenny Lau
19.9k2159
$endgroup$
$$exp(x) = 1 + int_0^{x} exp(t) mathrm dt > 1 + int_0^x 1 mathrm dt = 1 + x$$
answered Dec 31 '18 at 13:09
Kenny LauKenny Lau
19.9k2159
answered Dec 31 '18 at 13:09
Kenny LauKenny Lau
19.9k2159
answered Dec 31 '18 at 13:09
Kenny LauKenny Lau
19.9k2159
answered Dec 31 '18 at 13:09
Kenny LauKenny Lau
19.9k2159
19.9k2159
$begingroup$
That's what i was looking for! Beautiful method,thanks for taking the time to share it with me i'm grateful.
$endgroup$
– Bénz Anas
Jan 1 at 10:51
add a comment |
$begingroup$
That's what i was looking for! Beautiful method,thanks for taking the time to share it with me i'm grateful.
$endgroup$
– Bénz Anas
Jan 1 at 10:51
$begingroup$
That's what i was looking for! Beautiful method,thanks for taking the time to share it with me i'm grateful.
$endgroup$
– Bénz Anas
Jan 1 at 10:51
$begingroup$
That's what i was looking for! Beautiful method,thanks for taking the time to share it with me i'm grateful.
$endgroup$
– Bénz Anas
Jan 1 at 10:51
add a comment |
$begingroup$
Defining $$f(x)=e^x-x-1$$ then $$f'(x)=e^x-1$$ and $$f''(x)=e^x>0$$ Can you proceed?
answered Dec 31 '18 at 13:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42865
$endgroup$
1
$begingroup$
Sometimes I wonder how much of the questions you read. OP wrote: "The only method i use is to consider a new function F, that realizes F(x)=exp(x)-x-1 then calculate the derivative then use its monotony ... So do you have any better method?"
$endgroup$
– Martin R
Dec 31 '18 at 18:50
add a comment |
$begingroup$
Defining $$f(x)=e^x-x-1$$ then $$f'(x)=e^x-1$$ and $$f''(x)=e^x>0$$ Can you proceed?
answered Dec 31 '18 at 13:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42865
$endgroup$
1
$begingroup$
Sometimes I wonder how much of the questions you read. OP wrote: "The only method i use is to consider a new function F, that realizes F(x)=exp(x)-x-1 then calculate the derivative then use its monotony ... So do you have any better method?"
$endgroup$
– Martin R
Dec 31 '18 at 18:50
add a comment |
$begingroup$
Defining $$f(x)=e^x-x-1$$ then $$f'(x)=e^x-1$$ and $$f''(x)=e^x>0$$ Can you proceed?
answered Dec 31 '18 at 13:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42865
$endgroup$
Defining $$f(x)=e^x-x-1$$ then $$f'(x)=e^x-1$$ and $$f''(x)=e^x>0$$ Can you proceed?
answered Dec 31 '18 at 13:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42865
answered Dec 31 '18 at 13:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42865
answered Dec 31 '18 at 13:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42865
answered Dec 31 '18 at 13:09
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42865
73.7k42865
1
$begingroup$
Sometimes I wonder how much of the questions you read. OP wrote: "The only method i use is to consider a new function F, that realizes F(x)=exp(x)-x-1 then calculate the derivative then use its monotony ... So do you have any better method?"
$endgroup$
– Martin R
Dec 31 '18 at 18:50
add a comment |
1
$begingroup$
Sometimes I wonder how much of the questions you read. OP wrote: "The only method i use is to consider a new function F, that realizes F(x)=exp(x)-x-1 then calculate the derivative then use its monotony ... So do you have any better method?"
$endgroup$
– Martin R
Dec 31 '18 at 18:50
1
1
$begingroup$
Sometimes I wonder how much of the questions you read. OP wrote: "The only method i use is to consider a new function F, that realizes F(x)=exp(x)-x-1 then calculate the derivative then use its monotony ... So do you have any better method?"
$endgroup$
– Martin R
Dec 31 '18 at 18:50
$begingroup$
Sometimes I wonder how much of the questions you read. OP wrote: "The only method i use is to consider a new function F, that realizes F(x)=exp(x)-x-1 then calculate the derivative then use its monotony ... So do you have any better method?"
$endgroup$
– Martin R
Dec 31 '18 at 18:50
add a comment |
