Does the function y=x^3+x+(x^2+1)e^x have a horizontal tangent?

We want to determine whether a function has a tangent which is horizontal. To answer the question we need to know what a horizontal tangent to a function means. A tangent to a graph of a function is a line which "touches" the graph at a point. This means that the slope of the tangent at that particular point is the same as the gradient of the function. But the gradient of a function at any point is given by the derivative of the function. A horizontal tangent is a tangent with constant y-value so the slope, the derivative, is zero. It does not change. Hence to solve the problem we need to investigate whether the derivative of the function, y', can take the value 0. The function y is a sum of three terms, and so we can use the sum rule in differentiation which says that the derivative of a sum is the sum of the derivatives of the terms. The derivative of x^3 is 3x^(3-1)=3x^2 and the derivative of x is 1. However, to differentiate (x^2+1)e^x we need the product rule. We get ((x^2+1)e^x)'=(x^2+1)'e^x+(x^2+1)(e^x)'=(2x)(e^x)+(x^2+1)e^x rembering that (e^x)'=e^x. Hence we have y'=3x^2+1+2xe^x+(x^2+1)e^x and rearranging we get y'=3x^2+1+e^x(x^2+2x+1). Ok, so now we know an expression for y'. We want to know whether y'=0 is possible. We note that 3x^2 is always nonnegative (a square is zero or positive) and e^x is always positive (any power is positive). We now recognize x^2+2x+1 as the square of (x+1), and so x^2+2x+1=(x+1)^2≥0. Therefore e^x( x^2+2x+1)≥00=0 also. So putting it all together we have y' = 3x^2+1+e^x(x^2+2x+1) > 0+1+0=1. Hence the derivative y'is always greater than 1 and can therefore not be zero. Hence y has no horizontal tangent, in fact all tangents have slopes greater than 1.