A curve is defined by the equation $y = \ln\left(\frac{\text{e}^x}{x^2 + 10}\right)$. Show that the curve has no turning points for all real values of $x$.

1. Simplify the equation of the curve using the laws of logarithms.

$y = \ln\left(\frac{\text{e}^x}{x^2 + 10}\right)$
$y = \ln(\text{e}^x) - \ln(x^2 + 10)$
$y = x - \ln(x^2 + 10)$

2. Differentiate $y$ with respect to $x$.

$\frac{dy}{dx} = 1 - \frac{1}{x^2 + 10} \cdot (2x)$
$\frac{dy}{dx} = 1 - \frac{2x}{x^2 + 10}$

3. Equate the derivative to zero to find turning points and form a quadratic equation.

For turning points, $\frac{dy}{dx} = 0$.
$1 - \frac{2x}{x^2 + 10} = 0$
$\frac{x^2 + 10 - 2x}{x^2 + 10} = 0$
$x^2 - 2x + 10 = 0$

4. Analyse the roots of the quadratic equation using the discriminant.

Discriminant of $x^2 - 2x + 10 = 0$:
$b^2 - 4ac = (-2)^2 - 4(1)(10)$
$= 4 - 40 = -36$
Since $-36 < 0$, the equation $x^2 - 2x + 10 = 0$ has no real roots.
Therefore, $\frac{dy}{dx} = 0$ has no real solutions, meaning the curve has no turning points for all real values of $x$.