Let f be the real valued differentiable function on R such that e−x f(x)=3e2+4e−x∫2x2t2+6t+5 dt ∀x∈R and let g(x)=f−1(x) then g/ /(3) is equal to _______ (where f−1(x) is the inverse function of f(x))

g(f(x)) = xSo ; g/(f(x)) . f/(x)=1g/(f(x))=−f/ /(x)(f/(x))3So ; g/(3)=123 and g/ /(3)=− 435(23)3

Let f be the real valued differentiable function on R such that e−x f(x)=3e2+4e−x∫2x2t2+6t+5 dt ∀x∈R and let g(x)=f−1(x) then g/ /(3) is equal to _______ (where f−1(x) is the inverse function of f(x))

g(f(x)) = xSo ; g/(f(x)) . f/(x)=1g/(f(x))=−f/ /(x)(f/(x))3So ; g/(3)=123 and g/ /(3)=− 435(23)3

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Let f be the real valued differentiable function on R such that $e^{- x} f (x) = \frac{3}{e^{2}} + 4 e^{- x} \int_{2}^{x} \sqrt{2 t^{2} + 6 t + 5} d t \forall x \in R$ and let $g (x) = f^{- 1} (x)$ then $g^{/ /} (3)$ is equal to _______ (where $f^{- 1} (x)$ is the inverse function of $f (x)$ )

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$\frac{- 47}{5 {(23)}^{2}}$

$\frac{- 43}{5 {(23)}^{2}}$

$\frac{- 47}{5 {(23)}^{3}}$

$\frac{- 43}{5 {(23)}^{3}}$

answer is A.

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Detailed Solution

g(f(x)) = x
So ; $g^{/} (f (x)) . f^{/} (x) = 1$
$g^{/} (f (x)) = - \frac{f^{/ /} (x)}{{(f^{/} (x))}^{3}}$

So ; $g^{/} (3) = \frac{1}{23}$ and $g^{/ /} (3) = \frac{- 43}{5 {(23)}^{3}}$

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Let f be the real valued differentiable function on R such that e−x f(x)=3e2+4e−x∫2x2t2+6t+5 dt ∀x∈R and let g(x)=f−1(x) then g/ /(3) is equal to _______ (where f−1(x) is the inverse function of f(x))

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Let f be the real valued differentiable function on R such that $e^{- x} f (x) = \frac{3}{e^{2}} + 4 e^{- x} \int_{2}^{x} \sqrt{2 t^{2} + 6 t + 5} d t \forall x \in R$ and let $g (x) = f^{- 1} (x)$ then $g^{/ /} (3)$ is equal to _______ (where $f^{- 1} (x)$ is the inverse function of $f (x)$ )