Question

求下列函数的的值域f（x）=x²-2x x属于（1,6）f（x）=x-1+根号下x+1f（x）=x²

Answer 1

f(x) = x² - 2x = (x - 1)² - 1

1 < x < 6

0 < x - 1 < 5

0 < (x - 1)² < 25

-1 < f(x) < 24

f(x) = x - 1 + √(x + 1) = [√(x + 1) + 1/2]² - 9/4

√(x + 1) + 1/2 ≥ 1/2

[√(x + 1) + 1/2]² ≥ 1/4

f(x) ≥ -2

f(x)=x²/x²+1+x+1/x²+2

请明确一下分子分母