Let g (x) be a linear function and f(x) = \ cases g(x), & x ≤ 0 ((1...

Let, $g\left(x\right) = ax + b$

Now function $f\left(x\right)$ in continuous at $x = 0$

$⇒ \lim_{x \to {0}^{+}}f\left(x\right) = f\left(0\right)$

$⇒ \lim_{x \to 0}\left(\dfrac{1 + x}{2 + x}\right)\dfrac{1}{x} = b$

$⇒ 0 = b$ 

$⇒ g\left(x\right) = ax$

Now, for $x > 0$

$f\left(x\right) = \left(\dfrac{1 + x}{2 + x}\right)\dfrac{1}{x}$

$⇒ logf\left(x\right) = \log \left(\dfrac{1 + x}{2 + x}\right)\dfrac{1}{x}$

$⇒ logf\left(x\right) = \dfrac{\log \left(1 + x\right) - \log \left(2 + x\right)}{x}$

$⇒ \dfrac{1}{f\left(x\right)} \times {f}^{'}\left(x\right) = \dfrac{x\left(\dfrac{1}{1 + x} - \dfrac{1}{2 + x}\right) - \left[\log \left(1 + x\right) - \log \left(2 + x\right)\right]}{{x}^{2}}$

$⇒ {f}^{'}\left(x\right) = f\left(x\right)\left\{\dfrac{x\left(\dfrac{1}{1 + x} - \dfrac{1}{2 + x}\right) - \left[\log \left(1 + x\right) - \log \left(2 + x\right)\right]}{{x}^{2}}\right\}$

$⇒ {f}^{'}\left(1\right) = f\left(1\right)\left\{\dfrac{\left(\dfrac{1}{2} - \dfrac{1}{3}\right) - \left[\log \left(2\right) - \log \left(3\right)\right]}{{1}^{2}}\right\}$

$⇒ {f}^{'}\left(1\right) = \dfrac{2}{3}\left\{\dfrac{1}{6} - \log \left(\dfrac{2}{3}\right)\right\}$

And, $f\left(- 1\right) = g\left(- 1\right) = - a$

$⇒ a = \dfrac{2}{3}\log \left(\dfrac{2}{3}\right) - \dfrac{1}{9}$

Now,  finding $g\left(3\right) = 3a + 0$

$⇒ g\left(3\right) = 2\log \left(\dfrac{2}{3}\right) - \dfrac{1}{3}$

$⇒ g\left(3\right) = \log \left(\dfrac{4}{9 ⋅ {e}^{\dfrac{1}{3}}}\right)$