Suppose f and g are differentiable functions such that xg ( f ( x )) f ′ ( g ( x )) g ′ ( x ) = f ( g ( x )) g ′ ( f ( x )) f ′ ( x ) for all x ∈ R . Moreover, f is non-negative, g is positive, and ∫

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Suppose f and g are differentiable functions such that xg ( f ( x )) f ′ ( g ( x )) g ′ ( x ) = f ( g ( x )) g ′ ( f ( x )) f ′ ( x ) for all x ∈ R . Moreover, f is non-negative, g is positive, and ∫ 0 x ​ f ( g ( t )) d t = 2 1 − e − 2 x ​ for all x ∈ R . If g ( f ( 0 )) = 1 , then g ( f ( 4 )) is

TestHub · Accepted Answer

Differentiating the integral identity gives $f (g (x)) = e^{- 2 x} > 0$ . Divide the functional equation by $f (g (x)) g (f (x))$ to obtain

$\frac{d}{d x} log g (f (x)) = x \frac{d}{d x} log f (g (x)) = - 2 x .$

Therefore $g (f (x)) = C e^{- x^{2}}$ . The initial condition gives $C = 1$ and $g (f (4)) = e^{- 16}$ .

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