# `x^3 + e^x = 0` Show that the equation has exactly one real root.

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### 1 Answer

Denote `f(x)=x^3+e^x.` f is continuous everywhere.

When x tends to `+oo` , f(x) tends to `+oo` , when x tends to `-oo,` f(x) tends to `-oo.` In another words, there are `x_1` such that `f(x_1) lt 0` (for example, `x_1=-1` ) and `x_2` such that `f(x_2)gt0` (for example, `x_2=0` ).

By the Intermediate Value Theorem there is at least one c such that f(c)=0.

Let's prove that such c is unique. Really, f(x) is increases strictly monotonically as a sum of two such functions. If we want, we can use the derivative,` f'(x)=3x^2 + e^x gt 0.` Therefore f takes any value only once, QED.