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Step-by-Step Solution
Step 1: Express the relation in terms of logarithms
We are given that for a twice differentiable function
$f:(0,\infty)\to (0,\infty)$,
$f(x+1) = x \, f(x).$
Define
$g(x) = \ln \bigl(f(x)\bigr).$
Then,
$$
\ln \bigl(f(x+1)\bigr)
\;=\; \ln (x \, f(x))
\;=\; \ln x \;+\; \ln \bigl(f(x)\bigr),
$$
which implies
$$
g(x+1) \;=\; \ln x \;+\; g(x).
$$
Step 2: Derive the discrete difference equations for the derivatives
From
$g(x+1) - g(x) = \ln x,$
differentiate both sides with respect to $x$. We get:
$$
g'(x+1) - g'(x) \;=\; \frac{1}{x}.
$$
Differentiate again to find an expression involving the second derivatives:
$$
g''(x+1) - g''(x) \;=\; -\frac{1}{x^2}.
$$
Step 3: Evaluate the successive differences for specific integer values
We want
$g''(5) - g''(1).$
To find this, we sum up the differences for $x = 1, 2, 3, 4$:
For $x=1$:
$$
g''(2) - g''(1) = -\frac{1}{1^2} = -1,
$$
for $x=2$:
$$
g''(3) - g''(2) = -\frac{1}{2^2} = -\frac{1}{4},
$$
for $x=3$:
$$
g''(4) - g''(3) = -\frac{1}{3^2} = -\frac{1}{9},
$$
for $x=4$:
$$
g''(5) - g''(4) = -\frac{1}{4^2} = -\frac{1}{16}.
$$
Add these four equations:
$$
\bigl(g''(2) - g''(1)\bigr)
\;+\; \bigl(g''(3) - g''(2)\bigr)
\;+\; \bigl(g''(4) - g''(3)\bigr)
\;+\; \bigl(g''(5) - g''(4)\bigr)
= - \left( 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} \right).
$$
Notice the left-hand side telescopes to
$g''(5) - g''(1).$
Step 4: Simplify the telescoping sum
We compute:
$$
1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16}
= 1 + 0.25 + 0.111\ldots + 0.0625
= \frac{144}{144} + \frac{36}{144} + \frac{16}{144} + \frac{9}{144}
= \frac{144 + 36 + 16 + 9}{144}
= \frac{205}{144}.
$$
Therefore,
$$
g''(5) - g''(1)
= -\frac{205}{144}.
$$
Step 5: Find the absolute value
The question asks for
$\bigl| g''(5) - g''(1) \bigr|.$
Thus,
$$
\Bigl| g''(5) - g''(1) \Bigr|
= \left| -\frac{205}{144} \right|
= \frac{205}{144}.
$$
Final Answer
The value of
$\bigl|g''(5) - g''(1)\bigr|$
is
$\dfrac{205}{144}.$
