If the function [f:(-,-1] (a, b]] defined by f(x)=e^x^3-3 x+1 is one - one and onto, then …

Question

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 :

Accepted Answer

$$\begin{aligned}
& f(x)=e^{x^3-3 x+1} \
& f^{\prime}(x)=e^{x^3-3 x+1} \cdot\left(3 x^2-3\right) \
& =e^{x^3-3 x+1} \cdot 3(x-1)(x+1)
\end{aligned}$$

For $$\mathrm{f}^{\prime}(\mathrm{x}) \geq 0$$

$$\therefore \mathrm{f}(\mathrm{x})$$ is increasing function

$$\begin{aligned}
& \therefore \mathrm{a}=\mathrm{e}^{-\infty}=0=\mathrm{f}(-\infty) \
& \mathrm{b}=\mathrm{e}^{-1+3+1}=\mathrm{e}^3=\mathrm{f}(-1) \
& \mathrm{P}(2 \mathrm{~b}+4, \mathrm{a}+2) \
& \therefore \mathrm{P}\left(2 \mathrm{e}^3+4,2\right)
\end{aligned}$$

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

Solutions · 1