Find the value of $frac{df^{-1}}{dx}$ of the following function
$begingroup$
Suppose that the differentiable function $y = f(x)$ has an inverse and that the graph of ƒ passes through the point $(2, 4)$ and has a slope of $frac13$ there. Find the value of $frac{df^{-1}}{dx}$ at $x = 4 $
The answer in the textbook says it's $3$ but I keep getting different answers everytime I attempt the question.
My solution:
Using $(2,4)$ and $m = frac13$ I got $y = frac13x + frac{10}3$
Plugging $x = 4$ into the equation, I get $(4, frac{14}3)$
Using $(f^{-1})'(b) = frac{1}{f'(a)}$ I get $frac19$, which is obviously different from the textbook's answer.
Can someone help push me in the right direction?
calculus derivatives
edited Dec 12 '18 at 2:06
Kemono Chen
3,1941844
asked Dec 12 '18 at 1:57
herondaleherondale
437
$endgroup$
add a comment |
$begingroup$
Suppose that the differentiable function $y = f(x)$ has an inverse and that the graph of ƒ passes through the point $(2, 4)$ and has a slope of $frac13$ there. Find the value of $frac{df^{-1}}{dx}$ at $x = 4 $
The answer in the textbook says it's $3$ but I keep getting different answers everytime I attempt the question.
My solution:
Using $(2,4)$ and $m = frac13$ I got $y = frac13x + frac{10}3$
Plugging $x = 4$ into the equation, I get $(4, frac{14}3)$
Using $(f^{-1})'(b) = frac{1}{f'(a)}$ I get $frac19$, which is obviously different from the textbook's answer.
Can someone help push me in the right direction?
calculus derivatives
edited Dec 12 '18 at 2:06
Kemono Chen
3,1941844
asked Dec 12 '18 at 1:57
herondaleherondale
437
$endgroup$
add a comment |
$begingroup$
Suppose that the differentiable function $y = f(x)$ has an inverse and that the graph of ƒ passes through the point $(2, 4)$ and has a slope of $frac13$ there. Find the value of $frac{df^{-1}}{dx}$ at $x = 4 $
The answer in the textbook says it's $3$ but I keep getting different answers everytime I attempt the question.
My solution:
Using $(2,4)$ and $m = frac13$ I got $y = frac13x + frac{10}3$
Plugging $x = 4$ into the equation, I get $(4, frac{14}3)$
Using $(f^{-1})'(b) = frac{1}{f'(a)}$ I get $frac19$, which is obviously different from the textbook's answer.
Can someone help push me in the right direction?
calculus derivatives
edited Dec 12 '18 at 2:06
Kemono Chen
3,1941844
asked Dec 12 '18 at 1:57
herondaleherondale
437
$endgroup$
Suppose that the differentiable function $y = f(x)$ has an inverse and that the graph of ƒ passes through the point $(2, 4)$ and has a slope of $frac13$ there. Find the value of $frac{df^{-1}}{dx}$ at $x = 4 $
The answer in the textbook says it's $3$ but I keep getting different answers everytime I attempt the question.
My solution:
Using $(2,4)$ and $m = frac13$ I got $y = frac13x + frac{10}3$
Plugging $x = 4$ into the equation, I get $(4, frac{14}3)$
Using $(f^{-1})'(b) = frac{1}{f'(a)}$ I get $frac19$, which is obviously different from the textbook's answer.
Can someone help push me in the right direction?
calculus derivatives
calculus derivatives
edited Dec 12 '18 at 2:06
Kemono Chen
3,1941844
asked Dec 12 '18 at 1:57
herondaleherondale
437
edited Dec 12 '18 at 2:06
Kemono Chen
3,1941844
asked Dec 12 '18 at 1:57
herondaleherondale
437
edited Dec 12 '18 at 2:06
Kemono Chen
3,1941844
edited Dec 12 '18 at 2:06
Kemono Chen
3,1941844
edited Dec 12 '18 at 2:06
Kemono Chen
3,1941844
3,1941844
asked Dec 12 '18 at 1:57
herondaleherondale
437
asked Dec 12 '18 at 1:57
herondaleherondale
437
asked Dec 12 '18 at 1:57
herondaleherondale
437
437
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $y=f(x)$ passes through the point $(2,4)$, we can say that $f(2)=4$.
Therefore, $f^{-1}(4)=2$. Now$$dfrac{df^{-1}}{dx}bigg|_{x=4}=dfrac{1}{dfrac{df}{dx}bigg|_{x=f^{-1}(4)=2}}=dfrac{1}{dfrac13}=3$$
So, the value of $dfrac{df^{-1}}{dx}$ at $x=4$ is $3$.
answered Dec 12 '18 at 5:03
Key FlexKey Flex
8,57761233
$endgroup$
add a comment |
$begingroup$
You don't need the equation of the tangent line. One simply applies the formula that
$$
(f^{-1})'(f(a))=frac{1}{f'(a)}
$$
where in the problem $a=2$, $f(a)=4$ and $f'(a)=frac13$.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $y=f(x)$ passes through the point $(2,4)$, we can say that $f(2)=4$.
Therefore, $f^{-1}(4)=2$. Now$$dfrac{df^{-1}}{dx}bigg|_{x=4}=dfrac{1}{dfrac{df}{dx}bigg|_{x=f^{-1}(4)=2}}=dfrac{1}{dfrac13}=3$$
So, the value of $dfrac{df^{-1}}{dx}$ at $x=4$ is $3$.
answered Dec 12 '18 at 5:03
Key FlexKey Flex
8,57761233
$endgroup$
add a comment |
$begingroup$
Since $y=f(x)$ passes through the point $(2,4)$, we can say that $f(2)=4$.
Therefore, $f^{-1}(4)=2$. Now$$dfrac{df^{-1}}{dx}bigg|_{x=4}=dfrac{1}{dfrac{df}{dx}bigg|_{x=f^{-1}(4)=2}}=dfrac{1}{dfrac13}=3$$
So, the value of $dfrac{df^{-1}}{dx}$ at $x=4$ is $3$.
answered Dec 12 '18 at 5:03
Key FlexKey Flex
8,57761233
$endgroup$
add a comment |
$begingroup$
Since $y=f(x)$ passes through the point $(2,4)$, we can say that $f(2)=4$.
Therefore, $f^{-1}(4)=2$. Now$$dfrac{df^{-1}}{dx}bigg|_{x=4}=dfrac{1}{dfrac{df}{dx}bigg|_{x=f^{-1}(4)=2}}=dfrac{1}{dfrac13}=3$$
So, the value of $dfrac{df^{-1}}{dx}$ at $x=4$ is $3$.
answered Dec 12 '18 at 5:03
Key FlexKey Flex
8,57761233
$endgroup$
Since $y=f(x)$ passes through the point $(2,4)$, we can say that $f(2)=4$.
Therefore, $f^{-1}(4)=2$. Now$$dfrac{df^{-1}}{dx}bigg|_{x=4}=dfrac{1}{dfrac{df}{dx}bigg|_{x=f^{-1}(4)=2}}=dfrac{1}{dfrac13}=3$$
So, the value of $dfrac{df^{-1}}{dx}$ at $x=4$ is $3$.
answered Dec 12 '18 at 5:03
Key FlexKey Flex
8,57761233
answered Dec 12 '18 at 5:03
Key FlexKey Flex
8,57761233
answered Dec 12 '18 at 5:03
Key FlexKey Flex
8,57761233
answered Dec 12 '18 at 5:03
Key FlexKey Flex
8,57761233
8,57761233
add a comment |
add a comment |
$begingroup$
You don't need the equation of the tangent line. One simply applies the formula that
$$
(f^{-1})'(f(a))=frac{1}{f'(a)}
$$
where in the problem $a=2$, $f(a)=4$ and $f'(a)=frac13$.
$endgroup$
add a comment |
$begingroup$
You don't need the equation of the tangent line. One simply applies the formula that
$$
(f^{-1})'(f(a))=frac{1}{f'(a)}
$$
where in the problem $a=2$, $f(a)=4$ and $f'(a)=frac13$.
$endgroup$
add a comment |
$begingroup$
You don't need the equation of the tangent line. One simply applies the formula that
$$
(f^{-1})'(f(a))=frac{1}{f'(a)}
$$
where in the problem $a=2$, $f(a)=4$ and $f'(a)=frac13$.
$endgroup$
You don't need the equation of the tangent line. One simply applies the formula that
$$
(f^{-1})'(f(a))=frac{1}{f'(a)}
$$
where in the problem $a=2$, $f(a)=4$ and $f'(a)=frac13$.
answered Dec 12 '18 at 2:11
user587192
add a comment |
add a comment |
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