Calculus question!

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April 24th 2008, 03:20 AM
nugiboy

Calculus question!

'The gradient of a curve is given by $(dy)/(dx) = x^2 - 6x$ . Find the set of values of x for which y is an increasing function of x.

Im pretty good at calculus, but i'm not sure what it means by an increasing function of x.

Any help?
April 24th 2008, 03:22 AM
topsquark

Quote:

Originally Posted by nugiboy

'The gradient of a curve is given by $(dy)/(dx) = x^2 - 6x$ . Find the set of values of x for which y is an increasing function of x.

Im pretty good at calculus, but i'm not sure what it means by an increasing function of x.

Any help?

What does the derivative of a function mean? Thus if it is positive/negative what does that mean about the function?

-Dan
April 24th 2008, 03:28 AM
nugiboy

Quote:

Originally Posted by topsquark

What does the derivative of a function mean? Thus if it is positive/negative what does that mean about the function?

-Dan

Ah right so i need to find the values of x for which the gradient is positive?
Do i just do this by trial and error substituting in x values, or is there something im missing?
April 24th 2008, 04:10 AM
Isomorphism

Quote:

Originally Posted by nugiboy

Ah right so i need to find the values of x for which the gradient is positive?
Do i just do this by trial and error substituting in x values, or is there something im missing?

You want $x^2 - 6x > 0$

If x < 0, then x(x-6) > 0
If 0 < x < 6, then x(x-6) < 0
If x > 6, then x(x-6) > 0

So if $x \in ( - \infty, 0) \bigcup (6,\infty)$ , then $\frac{dy}{dx}$ is positive.

Or in English, except for values of x between 0 and 6, everywhere it is increasing :D
April 24th 2008, 04:16 AM
Moo

Hello,

A table of signs can help :

$x \leq 0 \text{ iff } x \leq 0$ (this can look unuseful, but it's the reasoning :))

$x-6 \leq 0 \text{ iff } x \leq 6$

So if $x \in ]-\infty,0] \ , \ x \leq 0 \text{ and } x-6 \leq 0. \text{ Hence } x(x-6) \geq 0$

If $x \in [0,6] \ , \ x \geq 0 \text{ and } x-6 \leq 0. \text{ Hence } x(x-6) \leq 0$

If $x \in [6,+\infty[ \ , \ x \geq 0 \text{ and } x-6 \geq 0. \text{ Hence } x(x-6) \geq 0$

So the function will be increasing for $\boxed{x \in ]- \infty,0] \cup [6,+\infty[} \ (= \mathbb{R} \backslash [0,6])$ :D