Answer to Question #320062 in Calculus for Grace

Question

Consider the curve y= f(x) for the function f(x)=e^2x-x^2

a)identify the domain of f

b)Find x,y-intercept

c)find the first and second derivatives

d)the critical points of the function f

e)where is the curve decreasing and increasing

F)find the point of inflection if any and discuss the concavity of the curve

G)Identify any asymptotes

Answer 1

$f\left( x \right) =e^{2x}-x^2\\a: Dom\left( f \right) =\mathbb{R} , because\,\,f\,\,is\,\,defined\,\,for\,\,all\,\,x\\b: f\left( 0 \right) =1-0=1\Rightarrow y-intercept\,\,is\,\,y=1\\f\left( x \right) =0\Rightarrow e^{2x}-x^2=0\Rightarrow x\approx -0.5671-this\,\,is\,\,x-intercept\\c:f'\left( x \right) =2e^{2x}-2x\\f''\left( x \right) =4e^{2x}-2\\d:\\f'\left( x \right) =0\Rightarrow 2e^{2x}-2x=0\Rightarrow e^{2x}=x\\No\,\,solutions\,\,for\,\,x\leqslant 0, \sin ce\,\,e^{2x}>0\\No\,\,solutions\,\,for\,\,x>0, \sin ce\,\,e^{2x}\geqslant 1+2x>x\\No\,\,critical\,\,points\\e: \\We\,\,have\,\,e^{2x}>x\,\,for\,\,all\,\,x\,\,from\,\,d\\Hence\,\,the\,\,curve\,\,is\,\,incre\mathrm{a}\sin g\,\,on\,\,\mathbb{R} \\f:\\f''\left( x \right) =0\Rightarrow 4e^{2x}-2=0\Rightarrow x=\frac{1}{2}\ln \frac{1}{2}=-\frac{\ln 2}{2}\\\left( -\infty ,-\frac{\ln 2}{2} \right) :f''<0-concave\,\,down\\\left( -\frac{\ln 2}{2},+\infty \right) :f''>0-concave\,\,up\\g:\\No\,\,vertical\,\,asymptotes\,\,because\,\,f\,\,in\,\,continuous\,\,on\,\,\mathbb{R} \\\underset{x\rightarrow +\infty}{\lim}\frac{f\left( x \right)}{x}=\underset{x\rightarrow +\infty}{\lim}\frac{e^{2x}-x^2}{x}=\left[ \frac{\infty}{\infty} \right] =\underset{x\rightarrow +\infty}{\lim}\frac{2e^{2x}-2x}{1}=+\infty \\\underset{x\rightarrow -\infty}{\lim}\frac{f\left( x \right)}{x}=\underset{x\rightarrow -\infty}{\lim}\frac{e^{2x}-x^2}{x}=\left[ \frac{-\infty}{-\infty} \right] =\underset{x\rightarrow -\infty}{\lim}\frac{2e^{2x}-2x}{1}=+\infty \\No\,\,asymptotes$