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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 realise 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 realise 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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