Inverse of function and derivative
0
$begingroup$
The problem says:
If $f(x)=frac{4x^3}{(x^2+1)}$ find $(f^{-1})'(2)$.
I can show that the function is one-to-one and maybe I should use $(f^{-1})'(x) = frac{1}{f'(f^{-1}(x))}$ but I dont know how.
The answer says (f^-1)'(2)=1/4
Thanks
calculus inverse
asked Dec 15 '18 at 12:37
L. G. L. G.
374
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add a comment |
0
$begingroup$
The problem says:
If $f(x)=frac{4x^3}{(x^2+1)}$ find $(f^{-1})'(2)$.
I can show that the function is one-to-one and maybe I should use $(f^{-1})'(x) = frac{1}{f'(f^{-1}(x))}$ but I dont know how.
The answer says (f^-1)'(2)=1/4
Thanks
calculus inverse
asked Dec 15 '18 at 12:37
L. G. L. G.
374
$endgroup$
add a comment |
0
0
0
1
$begingroup$
The problem says:
If $f(x)=frac{4x^3}{(x^2+1)}$ find $(f^{-1})'(2)$.
I can show that the function is one-to-one and maybe I should use $(f^{-1})'(x) = frac{1}{f'(f^{-1}(x))}$ but I dont know how.
The answer says (f^-1)'(2)=1/4
Thanks
calculus inverse
asked Dec 15 '18 at 12:37
L. G. L. G.
374
$endgroup$
The problem says:
If $f(x)=frac{4x^3}{(x^2+1)}$ find $(f^{-1})'(2)$.
I can show that the function is one-to-one and maybe I should use $(f^{-1})'(x) = frac{1}{f'(f^{-1}(x))}$ but I dont know how.
The answer says (f^-1)'(2)=1/4
Thanks
calculus inverse
calculus inverse
asked Dec 15 '18 at 12:37
L. G. L. G.
374
asked Dec 15 '18 at 12:37
L. G. L. G.
374
edited Dec 15 '18 at 13:10
L. G.
asked Dec 15 '18 at 12:37
L. G. L. G.
374
asked Dec 15 '18 at 12:37
L. G. L. G.
374
asked Dec 15 '18 at 12:37
L. G. L. G.
374
374
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint: Since $f(1)=2$, $f^{-1}(2)=1$.
answered Dec 15 '18 at 12:45
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
$begingroup$
ok, but how did you solve 2=f(x) for x?
$endgroup$
– L. G.
Dec 15 '18 at 12:47
$begingroup$
$$frac{4x^3}{x^2+1}=2iff4x^3=2x^2+2.$$Now, use the rational root theorem.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 12:49
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Hint: Since $f(1)=2$, $f^{-1}(2)=1$.
answered Dec 15 '18 at 12:45
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
$begingroup$
ok, but how did you solve 2=f(x) for x?
$endgroup$
– L. G.
Dec 15 '18 at 12:47
$begingroup$
$$frac{4x^3}{x^2+1}=2iff4x^3=2x^2+2.$$Now, use the rational root theorem.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 12:49
add a comment |
1
$begingroup$
Hint: Since $f(1)=2$, $f^{-1}(2)=1$.
answered Dec 15 '18 at 12:45
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
$begingroup$
ok, but how did you solve 2=f(x) for x?
$endgroup$
– L. G.
Dec 15 '18 at 12:47
$begingroup$
$$frac{4x^3}{x^2+1}=2iff4x^3=2x^2+2.$$Now, use the rational root theorem.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 12:49
add a comment |
1
1
1
$begingroup$
Hint: Since $f(1)=2$, $f^{-1}(2)=1$.
answered Dec 15 '18 at 12:45
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
Hint: Since $f(1)=2$, $f^{-1}(2)=1$.
answered Dec 15 '18 at 12:45
José Carlos SantosJosé Carlos Santos
160k22127232
answered Dec 15 '18 at 12:45
José Carlos SantosJosé Carlos Santos
160k22127232
answered Dec 15 '18 at 12:45
José Carlos SantosJosé Carlos Santos
160k22127232
answered Dec 15 '18 at 12:45
José Carlos SantosJosé Carlos Santos
160k22127232
160k22127232
$begingroup$
ok, but how did you solve 2=f(x) for x?
$endgroup$
– L. G.
Dec 15 '18 at 12:47
$begingroup$
$$frac{4x^3}{x^2+1}=2iff4x^3=2x^2+2.$$Now, use the rational root theorem.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 12:49
add a comment |
$begingroup$
ok, but how did you solve 2=f(x) for x?
$endgroup$
– L. G.
Dec 15 '18 at 12:47
$begingroup$
$$frac{4x^3}{x^2+1}=2iff4x^3=2x^2+2.$$Now, use the rational root theorem.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 12:49
$begingroup$
ok, but how did you solve 2=f(x) for x?
$endgroup$
– L. G.
Dec 15 '18 at 12:47
$begingroup$
ok, but how did you solve 2=f(x) for x?
$endgroup$
– L. G.
Dec 15 '18 at 12:47
$begingroup$
$$frac{4x^3}{x^2+1}=2iff4x^3=2x^2+2.$$Now, use the rational root theorem.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 12:49
$begingroup$
$$frac{4x^3}{x^2+1}=2iff4x^3=2x^2+2.$$Now, use the rational root theorem.
$endgroup$
– José Carlos Santos
Dec 15 '18 at 12:49
add a comment |
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