Find the function f (x) = X ²+ Definition field and value field of 5x + 49 / X

Find the function f (x) = X ²+ Definition field and value field of 5x + 49 / X

Answer:
f(x)=(x^2+5x+49)/x
The definition domain is x ≠ 0, i.e. (- ∞, 0) ∪ (0, + ∞)
f(x)=x+49/x+5
When x > 0, f (x) = x + 49 / x + 5 > = 2 √ [x * (49 / x)] + 5 = 2 * 7 + 5 = 19
When x

Function y = x under the root sign ²+ The domain of 5x-6 is___

x ²+ 5x-6 > = 0. X > = 1 or X

If the domain of function f (x) is [a, b], and b > - a > 0, the domain of function g (x) = f (x) - f (- x) is I know the answer is [a, - A]: take the intersection of F (x) and f (- x) domains. But what about the minus sign in the middle!

If you define a domain, just take the condition that x meets, and the minus sign in the middle can ignore it, because it does not affect the domain
However, if it is a division sign, it should be noted that the denominator cannot be 0

If the domain of function y = f (x) is [0,2], find the domain of function f (x2 + 2) It's hard to learn function just now! (⊙ o ⊙) I don't understand! My idea is, Because the definition field of function y = f (x) is [0,2], So 0

The arguments in F () brackets are the same domain. No matter how complex it is, just replace them directly

Let f (x) be a function defined on R, for m, n belong to R constant f (M + n) = f (m) * f (n), and when x > 0, 01, find the range of X

1. Let m = n, then f (2m) = F ² (M / 2) "0, so f (x) = f ² (x / 2) is nonnegative on the real number. Let n = 0, M > 0, then f (M + 0) = f (m) f (0), from which we can get f (0) = 1, let m = - n > 0, - M0, 0.2, let n be an infinitesimal positive real number, then M + n is slightly greater than m, then f (M + n) / F (m) = f (m) f (n) / F (m) = f

The function f (x) defined on (0, + ∞) has f (MN) = f (m) + F (n) for any m, n ∈ (0, + ∞). When x > 1, f (x) < 0 Judge and prove the monotonicity of F (x) in the definition domain 2. When f (2) = 1 / 2, solve the inequality f (AX + 4) > 1

1、 F (1) = 2F (1), f (1) = 0, let m > n > 0, then M / N > 1, f (n) + F (1 / N) = f (1) = 0, f (m) - f (n) = f (m) + F (1 / N) = f (M / N) < 0, so the function f (x) is a strictly monotonic decreasing function in the definition field (0, + ∞). Second, f (2) = 1 / 2, then f (4) = 1, so at 00, then - 4 / A