How do you find points of inflection and determine the intervals of concavity given #y=xe^(1/x)#?

Question

Emilia Aguilar · Accepted Answer

The function #y=xe^(1/x)# is convex for #x>0# and concave for #x<0#. It has no inflection points. It has discontinuity at #x=0# What does it mean? To determine concavity and points of inflection we need to find second derivative. If #y# is continuous at some point and: #y''>0<=>#function is convex (or concave up) #y''<0<=>#function is concave (or concave down) #y''=0# and it changes sign#<=>#inflection point. We will discover various features of this function and sketch the curve as we go. If you want to see how I got any of the following limit let me know in the comments. Before deriving #y=xe^(1/x)# Domain: #x in RR"\"{0}# or #x!=0# - no y intercept #y!=0# for #x!=0# - no x intercept Finding limits: #lim_(x->0^-)xe^(1/x)=0# #lim_(x->0^+)xe^(1/x)=+oo# #lim_(x->0^-)y!=lim_(x->0^+)y# so the limit at 0 doesn't exist and we have discontinuity. Assuming #x!=0# from now on. #lim_(x->+-oo)xe^(1/x)=+-oo# Because these limits are infinite, there could be up to 2 diagonal asymptotes #y=ax+b#. We find them by formulas #a=lim_(x->+-oo)y/x=lim_(x->+-oo)e^(1/x)=1# and #b=lim_(x->+-oo)y-ax=lim_(x->+-oo)xe^(1/x)-x=1#. It appears that both asymptotes exist and are the same. First derivative #y=xe^(1/x)# #y'=e^(1/x)+xe^(1/x)*(-1)/x^2# (product rule and chain rule) #y'=(1-1/x)e^(1/x)# Setting it to 0 to find extrema #(1-1/x)e^(1/x)=0# #(1-1/x)=0# #1=1/x# #x=1# - candidate for extremum. It must be a minimum, because here #y'# changes sign from negative to positive. Most limits of first derivative are already given from general function behavior. #lim_(x->0^+)dy/dx=-oo# #lim_(x->+-oo)dy/dx=1# One left to calculate #lim_(x->0^-)(x-1)/xe^(1/x)=0# Second derivative #y'=(1-1/x)e^(1/x)# #y''=1/x^2e^(1/x)+(1-1/x)e^(1/x)*(-1)/x^2# (product rule and chain rule) #y''=1/x^2e^(1/x)(1-(1-1/x))# #y''=e^(1/x)/x^3!=0# (thus no inflection points) Looking at sign we see that #y''# has the same sign as #x#, so your function #y=xe^(1/x)# is convex for #x>0# and concave for #x<0#. Graph of #xe^(1/x)# together with its diagonal asymptote: graph{(y-xe^(1/x))(y-x-1)sqrt(|x|-0.03)=0 [-10, 10, -5, 5]}

Aurora Alvarado · Answer

undefined To find points of inflection and determine intervals of concavity for the function $y = xe^{\frac{1}{x}}$, follow these steps: 1. **Find the first derivative $y'$** to understand the slope behavior: $$y' = e^{\frac{1}{x}} + x\left(-e^{\frac{1}{x}}\frac{1}{x^2}\right) = e^{\frac{1}{x}}\left(1 - \frac{1}{x^2}\right)$$ 2. **Find the second derivative $y''$** to analyze concavity: $$y'' = \frac{d}{dx}\left(e^{\frac{1}{x}}\left(1 - \frac{1}{x^2}\right)\right)$$ $$y'' = e^{\frac{1}{x}}\left(-\frac{1}{x^2}\right) + e^{\frac{1}{x}}\left(\frac{2}{x^3}\right)$$ $$y'' = e^{\frac{1}{x}}\left(\frac{2}{x^3} - \frac{1}{x^2}\right)$$ 3. **Solve $y'' = 0$ to find potential points of inflection**: $$e^{\frac{1}{x}}\left(\frac{2}{x^3} - \frac{1}{x^2}\right) = 0$$ Since $e^{\frac{1}{x}}$ is never 0, set the factor in parentheses equal to 0: $$\frac{2}{x^3} - \frac{1}{x^2} = 0$$ $$\frac{2 - x}{x^3} = 0$$ Solve for $x$: $$2 - x = 0$$ $$x = 2$$ 4. **Determine intervals of concavity**: Plug values of $x$ into $y''$ that are less than, between, and greater than the critical points found (in this case, $x = 2$) to determine the sign of $y''$. - For $x < 2$, use $x = 1$: $$y'' = e^{1}\left(\frac{2}{1^3} - \frac{1}{1^2}\right) = e(2 - 1) = e > 0$$ - For $x > 2$, use $x = 3$: $$y'' = e^{\frac{1}{3}}\left(\frac{2}{27} - \frac{1}{9}\right) = e^{\frac{1}{3}}\left(\frac{2 - 3}{27}\right) = -\frac{1}{27}e^{\frac{1}{3}} < 0$$ Hence, the function is concave up when $x < 2$ and concave down when $x > 2$. The point $x = 2$ is a point of inflection since the concavity changes from up to down as $x$ increases through $2$.