MATH HELP PLEASE!!!!!!!!ON LIMITS look hard but might be easy?

Question

Ok how do u show the equation x^3-cos^2x=0 has the root in [0,2]??????

and given that f(x)=x^4-x^2+5x+2, show that there exists at least two real numbers x such that f(x)=3.

Answer 1

Let g(x) = x^3-cos^2x. The statement says the root, so this suggests there's one, and only one, root in R.

g is continuous and g(0) = -1 < 0 and g(2) = 8 - cos^2(2) > 0, for cos(x) is in [-1, 1]. So, g has a root in (0,2). For x >2, g(x) > 8 - cos^2(x) > 0, so that g has no root >2. And if x < 0, x^3 <0 and g(x) < - cos^2(x) <=0, so tht g(x) <0. So, g has no negative rott. We conclude g has a unique root in (0,2), so in [0,2].

We have f(x) -3 = x^4-x^2+5x+-1. f -3 is a polynomial, so it's continuous. Also, f(0) = -1 < 0 and f(x) --> oo as x --> oo or x --> -oo (because f- 3 has even degree, 4). So, f-3 takes on positive values in (1, oo) and in (-oo, 1). Since it's continuous , it has a root in each of these intervals, showing it has at least 2 roots. This is the same as to say there exists at least two real numbers x such that f(x)=3.

Answer 2

f(x) = x^3 -cos^2(x)
f(0) = -1 < 0
f(2)= 8 - cos^2(2) > 0