# Show that a function is increasing

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.

## Homework Statement

Show that $f(x) = 9x^2 + 3x$ is strictly increasing on the interval $(0, 10]$

## 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 $u < v$ be two elements of the interval $(0, 10]$. Then from $f(u) < f(v)$ we obtain:
$$9u^2 + 3u < 9v^2 + 3v$$,
$$9u^2 + 3u - 9v^2 + 3v < 0$$

From here, I think I need to get to $u < v$, but I can't see how to do it.

Thanks for any help.

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arildno
Homework Helper
Gold Member
Dearly Missed
What sign does f(v)-f(u) have?

One thing you can do is evaluate the first derivative of the function on that specified interval.
Do you know how can you interpret the sign of the derivative of a function over a specified interval on how the function itself is behaving on that interval?

The sign must be $f(v) - f(u) > 0$, otherwise $f(v) < f(u)$.

Also, thanks for the suggestion about using the derivative, but I'm wondering if this can be done without.

Now for another go.

Let $u < v$ be in the interval $(0, 10]$. Suppose that $f(v) > f(u)$. Then:
$$f(v) - f(u) < 0$$
$$9v^2 + 3v - 9u^2 + 3u < 0$$
Which can only be true for $u > v$, hence, a contradiction.

Is that it, or did I miss a step?

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Dick
Homework Helper
Your reasoning is circular. If you really want to show its increasing without derivatives show f(v)-f(u) is positive by factoring the expression for it. Hint: it's divisible by (v-u). Correct the sign on the 3u term first.

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Parthalan - read carfully what Dick is suggesting.

Let u, v be two elelements in the interval (0,10] with u<v.

f(v)-f(u) = ................

and then as Dick says use the information you have to show this must be >0

I'm sure I have you all banging your head against a wall by now.

I got this far:

Let $f(x) = 9x^2 + 3x$ and $u, v \in (0, 10]$, where $u < v$. If the function is strictly increasing, then $f(u) < f(v)$. From this, we determine:
$$f(v) - f(u) = (9v^2 + 3v) - (9u^2 + 3u)$$
$$= 9v^2 + 3v - 9u^2 - 3u$$
$$= (-3u + 3v)(1 + 3u + 3v)$$
$$= -3(u - v)(1 + 3u + 3v)$$
Since $u < v$, then $u - v$ must be negative. Since the product of two negatives is always a positive, and $(1 + 3u + 3v)$ will also be positive, then $f(v) - f(u)$ must be positive, and therefore, $f(u) < f(v)$.
QED.

Is that right, or am I still going in circles?

Many thanks to the people who have helped.

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If $$0< x_{1} < x_{2}$$

$$3x_{1} < 3x_{2}[/itex] and [tex]9{x_{1}}^2 < 9{x_{2}}^2$$ . Therefore

$$9{x_{1}}^2 + 3x_{1} < 9{x_{2}}^2 + 3x_{2}$$. We're done here.

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Thanks!

That was what I tried to do in the first place, but I forgot to factor it first, and ended up trying to show that it was increasing because the difference between $f(x_2)$ and $f(x_1)$ was positive. Changing the inequality directly seems to be the easier way (not as easy as using the derivative, but unfortunately, it's a course that comes before calculus).

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One thing you can do is evaluate the first derivative of the function on that specified interval.
Do you know how can you interpret the sign of the derivative of a function over a specified interval on how the function itself is behaving on that interval?

Don't forget, this is the pre-calculus section! :tongue: