How do I find the inverse of a derivative at $x=-1$?
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Let $f(x)=x^3-3x^2-1$, $xgeq2$. Find the value of $left(df/dxright)^{-1}$ at $x=-1$.
This is the work that I've done so far:
Found $f'(x)=3x^2-3x$
Found the inverse of $f'(x)$ ►$3y^2-6y-x=0$
Found the roots of $y$ ►$[6±(36+12x)^{1/2}]/6$
Plugged $x=-1$ into $[6±(36+12x)^{1/2}]/6]$ and got $[3±(6)^{1/2}]/3$
Is $[3±(6)^{1/2}]/3$ the correct answer?
calculus inverse-function
edited Nov 26 at 16:57
amWhy
191k27223439
asked Nov 26 at 16:47
user597553
133
add a comment |
up vote
1
down vote
favorite
Let $f(x)=x^3-3x^2-1$, $xgeq2$. Find the value of $left(df/dxright)^{-1}$ at $x=-1$.
This is the work that I've done so far:
Found $f'(x)=3x^2-3x$
Found the inverse of $f'(x)$ ►$3y^2-6y-x=0$
Found the roots of $y$ ►$[6±(36+12x)^{1/2}]/6$
Plugged $x=-1$ into $[6±(36+12x)^{1/2}]/6]$ and got $[3±(6)^{1/2}]/3$
Is $[3±(6)^{1/2}]/3$ the correct answer?
calculus inverse-function
edited Nov 26 at 16:57
amWhy
191k27223439
asked Nov 26 at 16:47
user597553
133
Derivative is incorrect.
– Yadati Kiran
Nov 26 at 16:52
You say you found the inverse, but you didn't. At least it's now shown in your question that you did.
– Git Gud
Nov 26 at 17:05
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f(x)=x^3-3x^2-1$, $xgeq2$. Find the value of $left(df/dxright)^{-1}$ at $x=-1$.
This is the work that I've done so far:
Found $f'(x)=3x^2-3x$
Found the inverse of $f'(x)$ ►$3y^2-6y-x=0$
Found the roots of $y$ ►$[6±(36+12x)^{1/2}]/6$
Plugged $x=-1$ into $[6±(36+12x)^{1/2}]/6]$ and got $[3±(6)^{1/2}]/3$
Is $[3±(6)^{1/2}]/3$ the correct answer?
calculus inverse-function
edited Nov 26 at 16:57
amWhy
191k27223439
asked Nov 26 at 16:47
user597553
133
Let $f(x)=x^3-3x^2-1$, $xgeq2$. Find the value of $left(df/dxright)^{-1}$ at $x=-1$.
This is the work that I've done so far:
Found $f'(x)=3x^2-3x$
Found the inverse of $f'(x)$ ►$3y^2-6y-x=0$
Found the roots of $y$ ►$[6±(36+12x)^{1/2}]/6$
Plugged $x=-1$ into $[6±(36+12x)^{1/2}]/6]$ and got $[3±(6)^{1/2}]/3$
Is $[3±(6)^{1/2}]/3$ the correct answer?
calculus inverse-function
calculus inverse-function
edited Nov 26 at 16:57
amWhy
191k27223439
asked Nov 26 at 16:47
user597553
133
edited Nov 26 at 16:57
amWhy
191k27223439
asked Nov 26 at 16:47
user597553
133
edited Nov 26 at 16:57
amWhy
191k27223439
edited Nov 26 at 16:57
amWhy
191k27223439
edited Nov 26 at 16:57
amWhy
191k27223439
191k27223439
asked Nov 26 at 16:47
user597553
133
asked Nov 26 at 16:47
user597553
133
asked Nov 26 at 16:47
user597553
133
133
Derivative is incorrect.
– Yadati Kiran
Nov 26 at 16:52
You say you found the inverse, but you didn't. At least it's now shown in your question that you did.
– Git Gud
Nov 26 at 17:05
add a comment |
Derivative is incorrect.
– Yadati Kiran
Nov 26 at 16:52
You say you found the inverse, but you didn't. At least it's now shown in your question that you did.
– Git Gud
Nov 26 at 17:05
Derivative is incorrect.
– Yadati Kiran
Nov 26 at 16:52
Derivative is incorrect.
– Yadati Kiran
Nov 26 at 16:52
You say you found the inverse, but you didn't. At least it's now shown in your question that you did.
– Git Gud
Nov 26 at 17:05
You say you found the inverse, but you didn't. At least it's now shown in your question that you did.
– Git Gud
Nov 26 at 17:05
add a comment |
1 Answer
1
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oldest
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0
down vote
$f(x)=x^3-3x^2-1$.
- $f'(x)=3x^2-6x$
- For $(f'(x))^{-1} $ complete the squares. $y=3x^2-6x+3-3=3(x-1)^2-3implies x=1+sqrt{dfrac {y+3}{3}},:yin[-3,infty)$
answered Nov 26 at 17:00
Yadati Kiran
1,259417
1
The question changed from $df^{-1}/dx$ to $(df/dx)^{-1}$. Thats why the earlier question seemed absurd.
– Yadati Kiran
Nov 26 at 17:09
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
$f(x)=x^3-3x^2-1$.
- $f'(x)=3x^2-6x$
- For $(f'(x))^{-1} $ complete the squares. $y=3x^2-6x+3-3=3(x-1)^2-3implies x=1+sqrt{dfrac {y+3}{3}},:yin[-3,infty)$
answered Nov 26 at 17:00
Yadati Kiran
1,259417
1
The question changed from $df^{-1}/dx$ to $(df/dx)^{-1}$. Thats why the earlier question seemed absurd.
– Yadati Kiran
Nov 26 at 17:09
add a comment |
up vote
0
down vote
$f(x)=x^3-3x^2-1$.
- $f'(x)=3x^2-6x$
- For $(f'(x))^{-1} $ complete the squares. $y=3x^2-6x+3-3=3(x-1)^2-3implies x=1+sqrt{dfrac {y+3}{3}},:yin[-3,infty)$
answered Nov 26 at 17:00
Yadati Kiran
1,259417
1
The question changed from $df^{-1}/dx$ to $(df/dx)^{-1}$. Thats why the earlier question seemed absurd.
– Yadati Kiran
Nov 26 at 17:09
add a comment |
up vote
0
down vote
up vote
0
down vote
$f(x)=x^3-3x^2-1$.
- $f'(x)=3x^2-6x$
- For $(f'(x))^{-1} $ complete the squares. $y=3x^2-6x+3-3=3(x-1)^2-3implies x=1+sqrt{dfrac {y+3}{3}},:yin[-3,infty)$
answered Nov 26 at 17:00
Yadati Kiran
1,259417
$f(x)=x^3-3x^2-1$.
- $f'(x)=3x^2-6x$
- For $(f'(x))^{-1} $ complete the squares. $y=3x^2-6x+3-3=3(x-1)^2-3implies x=1+sqrt{dfrac {y+3}{3}},:yin[-3,infty)$
answered Nov 26 at 17:00
Yadati Kiran
1,259417
edited Nov 26 at 17:06
answered Nov 26 at 17:00
Yadati Kiran
1,259417
answered Nov 26 at 17:00
Yadati Kiran
1,259417
answered Nov 26 at 17:00
Yadati Kiran
1,259417
1,259417
1
The question changed from $df^{-1}/dx$ to $(df/dx)^{-1}$. Thats why the earlier question seemed absurd.
– Yadati Kiran
Nov 26 at 17:09
add a comment |
1
The question changed from $df^{-1}/dx$ to $(df/dx)^{-1}$. Thats why the earlier question seemed absurd.
– Yadati Kiran
Nov 26 at 17:09
1
1
The question changed from $df^{-1}/dx$ to $(df/dx)^{-1}$. Thats why the earlier question seemed absurd.
– Yadati Kiran
Nov 26 at 17:09
The question changed from $df^{-1}/dx$ to $(df/dx)^{-1}$. Thats why the earlier question seemed absurd.
– Yadati Kiran
Nov 26 at 17:09
add a comment |
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Derivative is incorrect.
– Yadati Kiran
Nov 26 at 16:52
You say you found the inverse, but you didn't. At least it's now shown in your question that you did.
– Git Gud
Nov 26 at 17:05