If the function f:(-infty,-1] rightarrow(a, b | vedclass

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(A) Given $f(x) = e^{x^3-3x+1}$ for $x \in (-\infty, -1]$.Find the derivative: $f'(x) = e^{x^3-3x+1} \cdot (3x^2-3) = 3e^{x^3-3x+1}(x-1)(x+1)$.For $x

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$2 \sqrt{1+e^6}$

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(A) Given $f(x) = e^{x^3-3x+1}$ for $x \in (-\infty, -1]$.Find the derivative: $f'(x) = e^{x^3-3x+1} \cdot (3x^2-3) = 3e^{x^3-3x+1}(x-1)(x+1)$.For $x \in (-\infty, -1]$,$x+1 \leq 0$ and $x-1 Thus,$f'(x) \geq 0$,meaning $f(x)$ is an increasing function on $(-\infty, -1]$.Since $f$ is one-one and onto,the range is $(a, b]$.$a = \lim_{x 	o -\infty} f(x) = e^{-\infty} = 0$.$b = f(-1) = e^{(-1)^3 - 3(-1) + 1} = e^{-1+3+1} = e^3$.So,$a=0$ and $b=e^3$.The point $P$ is $(2b+4, a+2) = (2e^3+4, 0+2) = (2e^3+4, 2)$.The distance $d$ of point $(x_1, y_1)$ from line $Ax+By+C=0$ is $d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$.Here,the line is $x + e^{-3}y - 4 = 0$,so $A=1, B=e^{-3}, C=-4$.$d = \frac{|1(2e^3+4) + e^{-3}(2) - 4|}{\sqrt{1^2 + (e^{-3})^2}} = \frac{|2e^3+4+2e^{-3}-4|}{\sqrt{1+e^{-6}}} = \frac{2e^3+2e^{-3}}{\sqrt{1+e^{-6}}}$.$d = \frac{2e^{-3}(e^6+1)}{\sqrt{\frac{e^6+1}{e^6}}} = \frac{2e^{-3}(e^6+1)}{\frac{\sqrt{e^6+1}}{e^3}} = 2\sqrt{e^6+1}$.

If the function $f:(-\infty,-1] \rightarrow(a, b]$ defined by $f(x)=e^{x^3-3 x+1}$ is one-one and onto,then the distance of the point $P(2 b+4, a+2)$ from the line $x+e^{-3} y=4$ is:

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