Let f : S rightarrow S where S =(0, infty) be | vedclass

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(A) Given $f(x+1) = xf(x)$. Taking the natural logarithm on both sides,we get $\ln f(x+1) = \ln(x f(x))$.Using the property $\ln(ab) = \ln a + \ln b$,

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$\frac{205}{144}$

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(A) Given $f(x+1) = xf(x)$. Taking the natural logarithm on both sides,we get $\ln f(x+1) = \ln(x f(x))$.Using the property $\ln(ab) = \ln a + \ln b$,we have $\ln f(x+1) = \ln x + \ln f(x)$.Since $g(x) = \ln f(x)$,this becomes $g(x+1) = \ln x + g(x)$,or $g(x+1) - g(x) = \ln x$.Differentiating both sides with respect to $x$ twice,we get $g''(x+1) - g''(x) = \frac{d^2}{dx^2}(\ln x) = -\frac{1}{x^2}$.Now,we substitute $x = 1, 2, 3, 4$ into this relation:For $x=1$: $g''(2) - g''(1) = -\frac{1}{1^2}$For $x=2$: $g''(3) - g''(2) = -\frac{1}{2^2}$For $x=3$: $g''(4) - g''(3) = -\frac{1}{3^2}$For $x=4$: $g''(5) - g''(4) = -\frac{1}{4^2}$Adding these four equations,the intermediate terms cancel out:$g''(5) - g''(1) = -\left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2}\right) = -\left(1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16}\right)$.Calculating the sum: $1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} = \frac{144 + 36 + 16 + 9}{144} = \frac{205}{144}$.Thus,$|g''(5) - g''(1)| = \frac{205}{144}$.

Let $f : S \rightarrow S$ where $S =(0, \infty)$ be a twice differentiable function such that $f(x+1) = xf(x)$. If $g : S \rightarrow R$ is defined as $g(x) = \log_{e} f(x)$,then the value of $|g''(5) - g''(1)|$ is equal to:

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