ok so i got the answer to that proble but can you verify it, Jay? x=-5,9 and interval notation is (-5,9) for some reason, my professor had the answer of the interval (-5,0)(9,infinity)
yes you and someone else are giving me different answers and its just screwing me up i found the zeros fine, its just the interval notation thats really pissing me off i plugged in 10 for the test point of the other interval and its not less than zero what the hell am i doing wrong
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i got more coming up
1/(t^2*t^(1/3)/t^4) * 1/(t^2*t^(1/3)/t^4) looks a lot easier to simplify because you know how multiplying exponents works.
well can you show me the steps please?
1*x^(-2/3) is obvious, x*x^(-2/3) = x^1*x^(-2/3) = x^(3/3)*x^(-2/3) = x^(1/3)
So you're left with x^(1/3) + x^(-2/3)
and the answer im supposed to be getting is
2x-1
x^2/3
That's how I got that x*x^(-2/3) = x^(1/3) part.
seeing soluations on here confuses me
dont worry about it
x=-5,9 and interval notation is (-5,9)
for some reason, my professor had the answer of the interval (-5,0)(9,infinity)
you and someone else are giving me different answers and its just screwing me up
i found the zeros fine, its just the interval notation thats really pissing me off
i plugged in 10 for the test point of the other interval and its not less than zero
what the hell am i doing wrong
45 - 4x - x^2 < 0
(9+x)(5-x) < 0
So x = -9 and 5
x can't be 0 because of the original problem.
By testing it you find that the interval from x = -9 to x = 0 is < 0 and the interval when x >5 is < 0
So the answer should be [-9,0) and [5, infinity].....
I realize that's not the answer you were given, but the graph I posted verifies it too and I don't think there are holes in the solution....