computing which function is above another function
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1
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Say we are trying to find the area between two functions.
So for example we have $y=e^x$ and $y=x^2-1$,$x=-1, x=1$. Graphing (or just knowing how these functions look) clearly shows $e^x$ is greater then $y=x^2-1$ within $[-1,1]$ but how would you figure that out in a mathematical way without graphing and seeing it visually?
calculus
asked Nov 13 at 20:45
user583753
618
|
show 5 more comments
up vote
1
down vote
favorite
Say we are trying to find the area between two functions.
So for example we have $y=e^x$ and $y=x^2-1$,$x=-1, x=1$. Graphing (or just knowing how these functions look) clearly shows $e^x$ is greater then $y=x^2-1$ within $[-1,1]$ but how would you figure that out in a mathematical way without graphing and seeing it visually?
calculus
asked Nov 13 at 20:45
user583753
618
Are you familiar with derivatives?
– Vasya
Nov 13 at 20:48
Yes. Covered those a long time ago.
– user583753
Nov 13 at 20:49
2
What about $e^x > 0 ge x^2 -1$ on that interval?
– Martin R
Nov 13 at 20:49
find derivative of $f(x)=e^x-x^2+1$ and show that it's positive on the interval
– Vasya
Nov 13 at 20:50
@Vasya: But of course if the derivative of $f(x)-g(x)$ is not everywhere positive in the interval, that does not mean that $f(x)$ is not always greater than $g(x)$. So your approach really isn't that helpful.
– David G. Stork
Nov 13 at 20:54
|
show 5 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Say we are trying to find the area between two functions.
So for example we have $y=e^x$ and $y=x^2-1$,$x=-1, x=1$. Graphing (or just knowing how these functions look) clearly shows $e^x$ is greater then $y=x^2-1$ within $[-1,1]$ but how would you figure that out in a mathematical way without graphing and seeing it visually?
calculus
asked Nov 13 at 20:45
user583753
618
Say we are trying to find the area between two functions.
So for example we have $y=e^x$ and $y=x^2-1$,$x=-1, x=1$. Graphing (or just knowing how these functions look) clearly shows $e^x$ is greater then $y=x^2-1$ within $[-1,1]$ but how would you figure that out in a mathematical way without graphing and seeing it visually?
calculus
calculus
asked Nov 13 at 20:45
user583753
618
asked Nov 13 at 20:45
user583753
618
asked Nov 13 at 20:45
user583753
618
asked Nov 13 at 20:45
user583753
618
asked Nov 13 at 20:45
user583753
618
618
Are you familiar with derivatives?
– Vasya
Nov 13 at 20:48
Yes. Covered those a long time ago.
– user583753
Nov 13 at 20:49
2
What about $e^x > 0 ge x^2 -1$ on that interval?
– Martin R
Nov 13 at 20:49
find derivative of $f(x)=e^x-x^2+1$ and show that it's positive on the interval
– Vasya
Nov 13 at 20:50
@Vasya: But of course if the derivative of $f(x)-g(x)$ is not everywhere positive in the interval, that does not mean that $f(x)$ is not always greater than $g(x)$. So your approach really isn't that helpful.
– David G. Stork
Nov 13 at 20:54
|
show 5 more comments
Are you familiar with derivatives?
– Vasya
Nov 13 at 20:48
Yes. Covered those a long time ago.
– user583753
Nov 13 at 20:49
2
What about $e^x > 0 ge x^2 -1$ on that interval?
– Martin R
Nov 13 at 20:49
find derivative of $f(x)=e^x-x^2+1$ and show that it's positive on the interval
– Vasya
Nov 13 at 20:50
@Vasya: But of course if the derivative of $f(x)-g(x)$ is not everywhere positive in the interval, that does not mean that $f(x)$ is not always greater than $g(x)$. So your approach really isn't that helpful.
– David G. Stork
Nov 13 at 20:54
Are you familiar with derivatives?
– Vasya
Nov 13 at 20:48
Are you familiar with derivatives?
– Vasya
Nov 13 at 20:48
Yes. Covered those a long time ago.
– user583753
Nov 13 at 20:49
Yes. Covered those a long time ago.
– user583753
Nov 13 at 20:49
2
2
What about $e^x > 0 ge x^2 -1$ on that interval?
– Martin R
Nov 13 at 20:49
What about $e^x > 0 ge x^2 -1$ on that interval?
– Martin R
Nov 13 at 20:49
find derivative of $f(x)=e^x-x^2+1$ and show that it's positive on the interval
– Vasya
Nov 13 at 20:50
find derivative of $f(x)=e^x-x^2+1$ and show that it's positive on the interval
– Vasya
Nov 13 at 20:50
@Vasya: But of course if the derivative of $f(x)-g(x)$ is not everywhere positive in the interval, that does not mean that $f(x)$ is not always greater than $g(x)$. So your approach really isn't that helpful.
– David G. Stork
Nov 13 at 20:54
@Vasya: But of course if the derivative of $f(x)-g(x)$ is not everywhere positive in the interval, that does not mean that $f(x)$ is not always greater than $g(x)$. So your approach really isn't that helpful.
– David G. Stork
Nov 13 at 20:54
|
show 5 more comments
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Are you familiar with derivatives?
– Vasya
Nov 13 at 20:48
Yes. Covered those a long time ago.
– user583753
Nov 13 at 20:49
2
What about $e^x > 0 ge x^2 -1$ on that interval?
– Martin R
Nov 13 at 20:49
find derivative of $f(x)=e^x-x^2+1$ and show that it's positive on the interval
– Vasya
Nov 13 at 20:50
@Vasya: But of course if the derivative of $f(x)-g(x)$ is not everywhere positive in the interval, that does not mean that $f(x)$ is not always greater than $g(x)$. So your approach really isn't that helpful.
– David G. Stork
Nov 13 at 20:54