Answer to Question #110661 in Calculus for please

Question

[img]https://upload.cc/i1/2020/04/18/6EeNfM.jpg[/img]


question in photo,R=4

Answer 1

a) $f(x)=(x-6)e^{-7x}$

i)

$f'(x)=e^{-7x}-7(x-6)e^{-7x}=e^{-7x}(43-7x)\\ f'(x)=0\\ e^{-7x}\neq0\quad 43-7x=0\\ x=\frac{43}{7}\\$

if $x\in(-\infty,\frac{43}{7}), f'(x)>0$ then $f$ is strictly increasing

if $x\in(\frac{43}{7},\infty), f'(x)<0$ then $f$ is strictly decreasing

ii)

$f''(x)=-7e^{-7x}(43-7x)-7e^{-7x}=\\ =-7e^{-7x}(44-7x)\\ f''(x)=0\\ x=\frac{44}{7}$

if $x\in (-\infty,\frac{44}{7}),f''(x)<0$ then $f$ is concave

if $x\in (\frac{44}{7},\infty), f''(x)>0$ then $f$ is convex

iii) $x=\frac{43}{7}$ is a point of max

$x=\frac{44}{7}$ is a point of inflexion

iiii)

the vertical asymptote does not exist

the horizontal asymptote

$\lim\limits_{x\to\infty}\frac{x-6}{e^{7x}}=0$

the horizontal asymptote $y=0$

the slant asymptote can be found in form

$y=kx+b$

let

$x\to+\infty\\ k=\lim\limits_{x\to+\infty}\frac{f(x)}{x}=\lim\limits_{x\to+\infty}\frac{(x-6)e^{-7x}}{x}=0\\ b=\lim\limits_{x\to+\infty}(f(x)-kx)=\lim\limits_{x\to+\infty}\frac{(x-6)}{e^{7x}}=0$

tnen $y=0$ is a slant asymptote

let

$x\to-\infty\\ k=\lim\limits_{x\to-\infty}\frac{f(x)}{x}=\lim\limits_{x\to-\infty}\frac{(x-6)e^{-7x}}{x}=\infty$

then a slant asymptote does not exist

b) $f=\frac{x^2+3x}{(x-1)^2}, x\neq1$

i)

$f'(x)=\frac{(2x+3)(x-1)^2-2(x-1)(x^2+3x)}{(x-1)^4}=\\ =\frac{-5x-3}{(x-1)^3}\\ f'(x)=0\\ x=-\frac{3}{5}$

if $x\in (-\infty,-\frac{3}{5})\cup(1,+\infty), f'(x)<0$ then $f$ is strictly decreasing

if $f\in(-\frac{3}{5},1), f'(x)>0$ then $f$ is strictly increasing

ii)

$f''=\frac{-5(x-1)^3-(x-1)^2(-5x-3)}{(x-1)^6}=\frac{10x+14}{(x-1)^4}\\ x=-\frac{7}{5}$

if $x\in(-\frac{7}{5},1)\cup(1,\infty), f''(x)>0$ then $f$ is convex

if $x\in(-\infty, -\frac{7}{5}), f''(x)<0$ then $f$ is concave

iii)

$x=-\frac{3}{5}$ is a point of min

$x=-\frac{7}{5}$ is a point of inflexion

iiii)

$x=1$ is the vertical asymptote

the horizontal asymptote

$\lim\limits_{x\to\infty}\frac{x^2+3x}{(x-1)^2}=1$

$y=1$ is a horizontal asymptote

the slant asymptote can be found in form

$y=kx+b$

$k=\lim\limits_{x\to\infty}\frac{f(x)}{x}=\lim\limits_{x\to\infty}\frac{x^2+3x}{x(x-1)^2}=0\\ b=\lim\limits_{x\to\infty}(f(x)-kx)=\lim\limits_{x\to\infty}\frac{x^2+3x}{(x-1)^2}=1$

tnen $y=1$ is a slant asymptote