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

Question

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

Accepted Answer

$\begin{gathered} f(x)={e}^{{x}^{3}-3x+1} \ {f}^{'}(x)={e}^{{x}^{3}-3x+1}\cdot (3{x}^{2}-3) \ ={e}^{{x}^{3}-3x+1}\cdot 3(x-1)(x+1) \end{gathered}$

For ${f}^{'}(x)\ge 0$

$	herefore f(x)$ is increasing function

$\begin{gathered} 	herefore a={e}^{-\infty}=0=f(-\infty) \ b={e}^{-1+3+1}={e}^{3}=f(-1) \ P(2b+4,a+2) \ 	herefore P(2{e}^{3}+4,2) \end{gathered}$

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

Solutions · 1