What are the local extrema of #f(x)= 1/sqrt(x^2+e^x)-xe^x#?

1 Answer
Jan 31, 2017


By graphical method, local maximum is 1.365, nearly, at the turning point (-0.555, 1.364), nearly. The curve has an asymptote #y = 0 larr#, the x-axis.


The approximations to the turning point (-0.555, 1.364), were obtained by moving lines parallel to the axes to meet at the zenith.

As indicated in the graph, it can be proved that, as #x to -oo, y to 0 and, as #x to oo, y to -oo#.

graph{(1/sqrt(x^2+e^x)-xe^x-y)(y-1.364)(x+.555+.001y)=0 [-10, 10, -5, 5]}