- #1
Parthalan
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Hi,
I'm in need of some help putting together this proof. I'm not sure if it's just harder than it looks, or I need to spend more time learning how to put them together.
Show that [itex]f(x) = 9x^2 + 3x[/itex] is strictly increasing on the interval [itex](0, 10][/itex]
I realize there are various arguments involving the turning point of a quadratic equation, but I don't think this is the type of answer that they're looking for.
Let [itex]u < v[/itex] be two elements of the interval [itex](0, 10][/itex]. Then from [itex]f(u) < f(v)[/itex] we obtain:
[tex]9u^2 + 3u < 9v^2 + 3v[/tex],
[tex]9u^2 + 3u - 9v^2 + 3v < 0[/tex]
From here, I think I need to get to [itex]u < v[/itex], but I can't see how to do it.
Thanks for any help.
I'm in need of some help putting together this proof. I'm not sure if it's just harder than it looks, or I need to spend more time learning how to put them together.
Homework Statement
Show that [itex]f(x) = 9x^2 + 3x[/itex] is strictly increasing on the interval [itex](0, 10][/itex]
Homework Equations
The Attempt at a Solution
I realize there are various arguments involving the turning point of a quadratic equation, but I don't think this is the type of answer that they're looking for.
Let [itex]u < v[/itex] be two elements of the interval [itex](0, 10][/itex]. Then from [itex]f(u) < f(v)[/itex] we obtain:
[tex]9u^2 + 3u < 9v^2 + 3v[/tex],
[tex]9u^2 + 3u - 9v^2 + 3v < 0[/tex]
From here, I think I need to get to [itex]u < v[/itex], but I can't see how to do it.
Thanks for any help.
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