Show that x^2 +2tan(x) -2 = 0 at some point?

f(0) = -2 < 0

f(pi/4) = (pi/4)^2 > 0

By continuity, f(x) = 0 at some point between 0 and pi/4.

Intermediate Value Theorem (IVT): If f is a continuous on the interval [a, b], and u is a number between f(a) and f(b), then there is a c ∈ [a, b] such that f(c) = u.

f(x) = x² + 2 tan(x) - 2

f(0) = -2 and f(π/4) = (π/4)²

-2 < 0 < (π/4)²

f is continuous on the interval [0,π/4], therefore by the IVT there is a c ∈ [0, π/4] such that f(c) = 0.

FYI, c ≈ 0.6626885692439121 in the interval [0,π/4]

Answer: see above

I'm sorry if i can't finished this answer..

x^2 + 2tan(x) - 2 = 0

x^2 - 2 = -2tanx

(x^2 - 2)/2 = -tanx

Uh, graph it and see that it has an x-intercept? An algebraic answer is another matter, but showing that it has a zero somewhere is fairly self-explanatory graphically speaking.