If the function f: (- ∞, - 1]→ (a, b)] defined by f(x) = e ^ x ^ 3...

Given: $f\left(x\right) = {e}^{{x}^{3} - 3x + 1}$

$⇒ {f}^{'}\left(x\right) = {e}^{{x}^{3} - 3x + 1} \times \left(3{x}^{2} - 3\right)$

$⇒ {f}^{'}\left(x\right) = 3 \times {e}^{{x}^{3} - 3x + 1} \times \left(x - 1\right) \times \left(x + 1\right)$

So, for $x \in ( - \infty , - 1 ]$ ${f}^{'}\left(x\right) > 0$, hence $f\left(x\right)$ is increasing function in given domain,

Now, finding $a$ and $b$ we get,

$⇒ a = \lim_{x \to - \infty}\left({e}^{{x}^{3} - 3x + 1}\right)$

$⇒ a = 0$

And $b = f\left(- 1\right)$

$⇒ b = {e}^{- 1 + 3 + 1}$

$⇒ b = {e}^{3}$

So, $P\left(2b + 4, a + 2\right)$

$⇒ P ≡ \left(2{e}^{3} + 4, 2\right)$

Now, finding the distance from the given line we get,

$⇒ d = \dfrac{\left(2{e}^{3} + 4\right) + 2{e}^{- 3} - 4}{\sqrt{1 + {e}^{- 6}}}$

$⇒ d = \dfrac{2\left({e}^{3} + {e}^{- 3}\right)}{\sqrt{1 + {e}^{- 6}}}$

$⇒ d = \dfrac{1}{{e}^{- 3}}\dfrac{2\left(1 + {e}^{- 6}\right)}{\sqrt{1 + {e}^{- 6}}}$

$⇒ d = \dfrac{2\sqrt{1 + {e}^{- 6}}}{{e}^{- 3}}$

$⇒ d = 2\sqrt{\dfrac{1}{{e}^{- 6}} + 1}$

$⇒ d = 2\sqrt{1 + {e}^{6}}$