Respuesta :
[tex](1\text{ , }\infty)[/tex]
Explanation:
Since for g(-∞), is not a possible answer. we remove it.
We also remove (∞,∞), because it goes back to being simply ∞.
Now for
(0 , 1 )
[tex]\begin{gathered} f(x)\text{ = x}^2\text{ - x } \\ f(1)\text{ = 1}^2\text{ - 1 = 0} \end{gathered}[/tex][tex]f(0)\text{ = 0}^2\text{ - 0 = 0}[/tex]This is not positive, nor negative so we will also put this one on the side.
For the last interval (1, ∞)
[tex]\begin{gathered} f(1)\text{ = 0 } \\ f(\infty)\text{ = }\infty^2-\infty\text{ = }\infty \end{gathered}[/tex]=> f(x) is positive
[tex]g(1)\text{ = log\lparen2*1 + 1\rparen = log\lparen3\rparen }\approx\text{ 0.477712}[/tex]g(1) is positive
[tex]g(\infty)\text{ = log\lparen2*}\infty+1)\text{ = log\lparen}\infty)\text{ = }\infty[/tex]=> g(∞) is positive
Hence f(x) and g(x) are both positve on the interval (1, 8)
NB:
Technically you should say g(x) tend to infinity (or a approaches infinity) not is equal to infinity, because infinity is not a number
