## Calculus 10th Edition

To find the slope of the tangents, we find the derivative first: $y'=(x^3+x)'=(x^3)'+(x)'=3x^2+1$ Horizontal means a slope of $0$ which indicates that the derivative should equal $0$: $y'=0\rightarrow3x^2+1=0\rightarrow x^2=-\frac{1}{3}$ Since over the domain of real numbers,$x^2$ cannot be negative then the above equation has no solution hence the derivative is never equal to $0$ and therefore there are no horizontal tangents.