How to find where a function is increasing at the greatest rate
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Given the function $f(x) = frac{1000x^2}{11+x^2}$ on the interval $[0, 3]$, how would I calculate where the function is increasing at the greatest rate?
Differentiating the function will give its slope. Since slope is defined as the rate of change, then getting the maxima of the function's derivative will indicate where it is increasing at the greatest rate.
The derivative of $f(x)$ is $frac{22000x}{(11+x^2)^2}$
Applying the first derivative test, the critical number is $sqrt{frac{11}{3}}$. The function increases before the critical number and decreases after it, so the critical number is a maximum. $sqrt{frac{11}{3}}$ is the answer.
calculus functions proof-verification
asked Dec 18 '15 at 4:27
hmir
16016
|
show 1 more comment
up vote
6
down vote
favorite
Given the function $f(x) = frac{1000x^2}{11+x^2}$ on the interval $[0, 3]$, how would I calculate where the function is increasing at the greatest rate?
Differentiating the function will give its slope. Since slope is defined as the rate of change, then getting the maxima of the function's derivative will indicate where it is increasing at the greatest rate.
The derivative of $f(x)$ is $frac{22000x}{(11+x^2)^2}$
Applying the first derivative test, the critical number is $sqrt{frac{11}{3}}$. The function increases before the critical number and decreases after it, so the critical number is a maximum. $sqrt{frac{11}{3}}$ is the answer.
calculus functions proof-verification
asked Dec 18 '15 at 4:27
hmir
16016
1
What have you tried? What do you know about calculating maxima? If you don't demonstrate any effort or context, we can't help you very effectively.
– Zach Stone
Dec 18 '15 at 4:30
"...so it is a maxima" - what is it in this statement? (And it should just be called a maximum, singular)
– The Chaz 2.0
Dec 18 '15 at 5:06
@The Chaz 2.0, Edited!
– hmir
Dec 18 '15 at 5:08
2
Right. "Where the function reaches its maximum on the interval" is not the same as "where the rate of change is maximized on the interval". You need to take the second derivative.
– The Chaz 2.0
Dec 18 '15 at 5:11
It is not without interest for the beginner to look at the following difference where the fastest rate of growth is noted shortly before $sqrt{frac{11}{3}}$: $$fleft(sqrt{dfrac{11}{3}+0.1}right)-fleft(sqrt{dfrac{11}{3}}right)approx5.07900\fleft(sqrt{dfrac{11}{3}}right)-fleft(sqrt{dfrac{11}{3}-0.1}right)approx5.14875$$
– Piquito
Apr 5 at 12:07
|
show 1 more comment
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Given the function $f(x) = frac{1000x^2}{11+x^2}$ on the interval $[0, 3]$, how would I calculate where the function is increasing at the greatest rate?
Differentiating the function will give its slope. Since slope is defined as the rate of change, then getting the maxima of the function's derivative will indicate where it is increasing at the greatest rate.
The derivative of $f(x)$ is $frac{22000x}{(11+x^2)^2}$
Applying the first derivative test, the critical number is $sqrt{frac{11}{3}}$. The function increases before the critical number and decreases after it, so the critical number is a maximum. $sqrt{frac{11}{3}}$ is the answer.
calculus functions proof-verification
asked Dec 18 '15 at 4:27
hmir
16016
Given the function $f(x) = frac{1000x^2}{11+x^2}$ on the interval $[0, 3]$, how would I calculate where the function is increasing at the greatest rate?
Differentiating the function will give its slope. Since slope is defined as the rate of change, then getting the maxima of the function's derivative will indicate where it is increasing at the greatest rate.
The derivative of $f(x)$ is $frac{22000x}{(11+x^2)^2}$
Applying the first derivative test, the critical number is $sqrt{frac{11}{3}}$. The function increases before the critical number and decreases after it, so the critical number is a maximum. $sqrt{frac{11}{3}}$ is the answer.
calculus functions proof-verification
calculus functions proof-verification
asked Dec 18 '15 at 4:27
hmir
16016
asked Dec 18 '15 at 4:27
hmir
16016
edited Dec 18 '15 at 5:08
asked Dec 18 '15 at 4:27
hmir
16016
asked Dec 18 '15 at 4:27
hmir
16016
asked Dec 18 '15 at 4:27
hmir
16016
16016
1
What have you tried? What do you know about calculating maxima? If you don't demonstrate any effort or context, we can't help you very effectively.
– Zach Stone
Dec 18 '15 at 4:30
"...so it is a maxima" - what is it in this statement? (And it should just be called a maximum, singular)
– The Chaz 2.0
Dec 18 '15 at 5:06
@The Chaz 2.0, Edited!
– hmir
Dec 18 '15 at 5:08
2
Right. "Where the function reaches its maximum on the interval" is not the same as "where the rate of change is maximized on the interval". You need to take the second derivative.
– The Chaz 2.0
Dec 18 '15 at 5:11
It is not without interest for the beginner to look at the following difference where the fastest rate of growth is noted shortly before $sqrt{frac{11}{3}}$: $$fleft(sqrt{dfrac{11}{3}+0.1}right)-fleft(sqrt{dfrac{11}{3}}right)approx5.07900\fleft(sqrt{dfrac{11}{3}}right)-fleft(sqrt{dfrac{11}{3}-0.1}right)approx5.14875$$
– Piquito
Apr 5 at 12:07
|
show 1 more comment
1
What have you tried? What do you know about calculating maxima? If you don't demonstrate any effort or context, we can't help you very effectively.
– Zach Stone
Dec 18 '15 at 4:30
"...so it is a maxima" - what is it in this statement? (And it should just be called a maximum, singular)
– The Chaz 2.0
Dec 18 '15 at 5:06
@The Chaz 2.0, Edited!
– hmir
Dec 18 '15 at 5:08
2
Right. "Where the function reaches its maximum on the interval" is not the same as "where the rate of change is maximized on the interval". You need to take the second derivative.
– The Chaz 2.0
Dec 18 '15 at 5:11
It is not without interest for the beginner to look at the following difference where the fastest rate of growth is noted shortly before $sqrt{frac{11}{3}}$: $$fleft(sqrt{dfrac{11}{3}+0.1}right)-fleft(sqrt{dfrac{11}{3}}right)approx5.07900\fleft(sqrt{dfrac{11}{3}}right)-fleft(sqrt{dfrac{11}{3}-0.1}right)approx5.14875$$
– Piquito
Apr 5 at 12:07
1
1
What have you tried? What do you know about calculating maxima? If you don't demonstrate any effort or context, we can't help you very effectively.
– Zach Stone
Dec 18 '15 at 4:30
What have you tried? What do you know about calculating maxima? If you don't demonstrate any effort or context, we can't help you very effectively.
– Zach Stone
Dec 18 '15 at 4:30
"...so it is a maxima" - what is it in this statement? (And it should just be called a maximum, singular)
– The Chaz 2.0
Dec 18 '15 at 5:06
"...so it is a maxima" - what is it in this statement? (And it should just be called a maximum, singular)
– The Chaz 2.0
Dec 18 '15 at 5:06
@The Chaz 2.0, Edited!
– hmir
Dec 18 '15 at 5:08
@The Chaz 2.0, Edited!
– hmir
Dec 18 '15 at 5:08
2
2
Right. "Where the function reaches its maximum on the interval" is not the same as "where the rate of change is maximized on the interval". You need to take the second derivative.
– The Chaz 2.0
Dec 18 '15 at 5:11
Right. "Where the function reaches its maximum on the interval" is not the same as "where the rate of change is maximized on the interval". You need to take the second derivative.
– The Chaz 2.0
Dec 18 '15 at 5:11
It is not without interest for the beginner to look at the following difference where the fastest rate of growth is noted shortly before $sqrt{frac{11}{3}}$: $$fleft(sqrt{dfrac{11}{3}+0.1}right)-fleft(sqrt{dfrac{11}{3}}right)approx5.07900\fleft(sqrt{dfrac{11}{3}}right)-fleft(sqrt{dfrac{11}{3}-0.1}right)approx5.14875$$
– Piquito
Apr 5 at 12:07
It is not without interest for the beginner to look at the following difference where the fastest rate of growth is noted shortly before $sqrt{frac{11}{3}}$: $$fleft(sqrt{dfrac{11}{3}+0.1}right)-fleft(sqrt{dfrac{11}{3}}right)approx5.07900\fleft(sqrt{dfrac{11}{3}}right)-fleft(sqrt{dfrac{11}{3}-0.1}right)approx5.14875$$
– Piquito
Apr 5 at 12:07
|
show 1 more comment
1 Answer
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Differentiating the function will give its slope. Since slope is defined as the rate of change, then getting the maxima of the function's derivative will indicate where it is increasing at the greatest rate.
The derivative of $f(x)$ is $frac{22000x}{(11+x^2)^2}$
Applying the first derivative test, the critical number is $sqrt{frac{11}{3}}$. The function increases before the critical number and decreases after it, so the critical number is a maximum. $sqrt{frac{11}{3}}$ is the answer.
answered Dec 18 '15 at 4:48
hmir
16016
1
This should go up above! I copied this answer and put it in the body of your question.
– graydad
Dec 18 '15 at 5:04
add a comment |
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Differentiating the function will give its slope. Since slope is defined as the rate of change, then getting the maxima of the function's derivative will indicate where it is increasing at the greatest rate.
The derivative of $f(x)$ is $frac{22000x}{(11+x^2)^2}$
Applying the first derivative test, the critical number is $sqrt{frac{11}{3}}$. The function increases before the critical number and decreases after it, so the critical number is a maximum. $sqrt{frac{11}{3}}$ is the answer.
answered Dec 18 '15 at 4:48
hmir
16016
1
This should go up above! I copied this answer and put it in the body of your question.
– graydad
Dec 18 '15 at 5:04
add a comment |
up vote
0
down vote
Differentiating the function will give its slope. Since slope is defined as the rate of change, then getting the maxima of the function's derivative will indicate where it is increasing at the greatest rate.
The derivative of $f(x)$ is $frac{22000x}{(11+x^2)^2}$
Applying the first derivative test, the critical number is $sqrt{frac{11}{3}}$. The function increases before the critical number and decreases after it, so the critical number is a maximum. $sqrt{frac{11}{3}}$ is the answer.
answered Dec 18 '15 at 4:48
hmir
16016
1
This should go up above! I copied this answer and put it in the body of your question.
– graydad
Dec 18 '15 at 5:04
add a comment |
up vote
0
down vote
up vote
0
down vote
Differentiating the function will give its slope. Since slope is defined as the rate of change, then getting the maxima of the function's derivative will indicate where it is increasing at the greatest rate.
The derivative of $f(x)$ is $frac{22000x}{(11+x^2)^2}$
Applying the first derivative test, the critical number is $sqrt{frac{11}{3}}$. The function increases before the critical number and decreases after it, so the critical number is a maximum. $sqrt{frac{11}{3}}$ is the answer.
answered Dec 18 '15 at 4:48
hmir
16016
Differentiating the function will give its slope. Since slope is defined as the rate of change, then getting the maxima of the function's derivative will indicate where it is increasing at the greatest rate.
The derivative of $f(x)$ is $frac{22000x}{(11+x^2)^2}$
Applying the first derivative test, the critical number is $sqrt{frac{11}{3}}$. The function increases before the critical number and decreases after it, so the critical number is a maximum. $sqrt{frac{11}{3}}$ is the answer.
answered Dec 18 '15 at 4:48
hmir
16016
edited Dec 18 '15 at 5:08
answered Dec 18 '15 at 4:48
hmir
16016
answered Dec 18 '15 at 4:48
hmir
16016
answered Dec 18 '15 at 4:48
hmir
16016
16016
1
This should go up above! I copied this answer and put it in the body of your question.
– graydad
Dec 18 '15 at 5:04
add a comment |
1
This should go up above! I copied this answer and put it in the body of your question.
– graydad
Dec 18 '15 at 5:04
1
1
This should go up above! I copied this answer and put it in the body of your question.
– graydad
Dec 18 '15 at 5:04
This should go up above! I copied this answer and put it in the body of your question.
– graydad
Dec 18 '15 at 5:04
add a comment |
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1
What have you tried? What do you know about calculating maxima? If you don't demonstrate any effort or context, we can't help you very effectively.
– Zach Stone
Dec 18 '15 at 4:30
"...so it is a maxima" - what is it in this statement? (And it should just be called a maximum, singular)
– The Chaz 2.0
Dec 18 '15 at 5:06
@The Chaz 2.0, Edited!
– hmir
Dec 18 '15 at 5:08
2
Right. "Where the function reaches its maximum on the interval" is not the same as "where the rate of change is maximized on the interval". You need to take the second derivative.
– The Chaz 2.0
Dec 18 '15 at 5:11
It is not without interest for the beginner to look at the following difference where the fastest rate of growth is noted shortly before $sqrt{frac{11}{3}}$: $$fleft(sqrt{dfrac{11}{3}+0.1}right)-fleft(sqrt{dfrac{11}{3}}right)approx5.07900\fleft(sqrt{dfrac{11}{3}}right)-fleft(sqrt{dfrac{11}{3}-0.1}right)approx5.14875$$
– Piquito
Apr 5 at 12:07