Question

Question #aae8e

Answer 1

See Below

Explanation:

Let `f(x) = x^3 - e^-x`

Then `f(0) = 0-e^0 = -1`
And, `f(1) = 1 - e^-1 = 0.6321` (4dp)

Both `e^-x` and `x^3` are continuous functions `AA x in RR`, and hence `f(x) = x^3 - e^-x` is a continuous function `AA x in RR`. As `f(x)` changes sign over the interval `[0,1]` then `f(x)=0` must have a root in that interval, and consequently `e^-x=x^3` has a solution in the interval.