Intermediate Value Theorem: ln x = e ^ -x (1, 2)?

Let f(x) = ln x - e^(-x) on [1,2].

Now f is continuous on [1, 2];

f(1) = 0 - e^(-1) = -1/e < 0

and f(2) = ln 2 - e^(-2) ≈ 0.558 > 0

So by the IVT there must be a c ∈ (1, 2) such that f(c) = 0 and hence ln c = e^(-c).

Ivt Theorem

i might want to start up by technique of subtracting the arctan time period to the different aspect to get ln(x) - arctan(x) = 0. Inverse tangent takes on values from -pi/2 to pi/2, so its insignificant for this situation. organic log takes on values from - infinity to infinity, so take a large unfavorable value and a large effectual value. because you'll discover such large extremums, by technique of the IVT, a nil ought to exist because that is a continuous function.