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

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

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

To find the local extrema of the function $f(x)=\frac{1}{\sqrt{x^2+e^x}}-xe^x$, we need to find the critical points where the derivative is either zero or undefined.

1. Find the derivative $f'(x)$.
2. Set $f'(x)=0$ and solve for $x$.
3. Check for points where $f'(x)$ is undefined.
4. Identify the local extrema among the critical points.

Let's start by finding the derivative:

$f(x) = \frac{1}{\sqrt{x^2+e^x}}-xe^x$

$f'(x) = \frac{d}{dx}\left(\frac{1}{\sqrt{x^2+e^x}}\right) - \frac{d}{dx}(xe^x)$

$f'(x) = -\frac{x e^x}{(e^x+x^2)^{\frac{3}{2}}} - e^x - xe^x$

Now, set $f'(x)=0$ and solve for $x$:

$-\frac{x e^x}{(e^x+x^2)^{\frac{3}{2}}} - e^x - xe^x = 0$

$-x e^x - (e^x)(e^x+x^2) - x(e^x)(e^x+x^2)^{\frac{3}{2}} = 0$

The critical points are where the derivative is either zero or undefined. However, finding an algebraic solution for $x$ in this case is complex and may require numerical methods.