Prove that y = 3x ^ 2 + 1 is a continuous function in the domain of definition by limit

Prove that y = 3x ^ 2 + 1 is a continuous function in the domain of definition by limit

Because u = x & # 178; is a continuous function in X ∈ (- infinity, + ∞)
So v = 3U = 3x & # 178; is a continuous function in X ∈ (- infinity, + ∞)
So y = V + 1 = 3x & # 178; + 1 is a continuous function in X ∈ (- infinity, + ∞)

Given that f (x) is an increasing function defined on (0, + ∞), and f (XY) = f (x) - f (y), f (2) = 1, the inequality f (x) - f (1x − 3) ≤ 2 is solved

∵ f (XY) = f (x) - f (y), ∵ f (XY) + F (y) = f (x), ∵ f (2) = 1, ∵ 2 = f (2) + F (2), let y = 2, xy = 2, that is, x = 2Y = 4, then f (2) + F (2) = f (4) = 2, then the inequality f (x) - f (1 x − 3) ≤ 2. It is equivalent to the inequality f [x (x-3)] ≤ f (4). ∵ f (x) is an increasing function defined on (0, + ∞), and the inequality is equivalent to X (x − 3) ≤ 4x 4 The solution is 3 ＜ x ≤ 4, that is, the solution set of the inequality is (3,4]

It is known that f (x) is an increasing function defined on (0, + ∞), and satisfies f (x? Y) = f (x) + F (y), f (2) = 1, and solves the inequality f (x) - f (X-2) > 3
Sorry, it's multiplication, that is, x times y

f(x.y)=f(x)+f(y),f(4)=2f(2)=2 f(8)=f(4)+f(2)=3
f(x)-f(x-2)>3 f(x)>f(x-2)+3=f(x-2)+f(8)=f(8x-16)
x>8x-16 x2
two