# Increasing and Decreasing Functions

• February 27th 2008, 12:05 PM
kdogg121
Increasing and Decreasing Functions
Hi, I am stuck on this problem, the instructions indicate to find the intervals of increase and decrease for the given function.

h(u) = square root of 9-u^2 The square root is over the entire term.

We were told to differentiate first, and then to factor to find the zeros. When I try to differentiate I am getting 3-(1/2)X^-(3/2) This doesn't look right so I need some clarification as to whether I am differentiating correctly.

BTW The final answer should be:

h(u) is increasing for -3< u < 0
h(u) is decreasing for 0< u < 3
• February 27th 2008, 12:13 PM
mr fantastic
Quote:

Originally Posted by kdogg121
Hi, I am stuck on this problem, the instructions indicate to find the intervals of increase and decrease for the given function.

h(u) = square root of 9-u^2 The square root is over the entire term.

We were told to differentiate first, and then to factor to find the zeros. When I try to differentiate I am getting 3-(1/2)X^-(3/2) This doesn't look right so I need some clarification as to whether I am differentiating correctly.

BTW The final answer should be:

h(u) is increasing for -3< u < 0
h(u) is decreasing for 0< u < 3

$h(u) = \sqrt{9 - u^2}$.

From the chain rule: $h^{'}(u) = \frac{1}{2} (9 - u^2)^{-1/2} \times (-2u) = -\frac{u}{\sqrt{9 - u^2}}$

where I've used the index laws $\sqrt{a} = a^{1/2}$ and $a^{-1/2} = \frac{1}{a^{1/2}} = \frac{1}{\sqrt{a}}$.

Important note: h(u) is only defined for $-3 \leq u \leq 3$ .....
• February 27th 2008, 12:18 PM
kdogg121
So was I correct then? 3-(1/2)U^-(1/2) is the derivative?

Will this factor?
• February 27th 2008, 02:08 PM
hatsoff
Quote:

Originally Posted by kdogg121
So was I correct then? 3-(1/2)U^-(1/2) is the derivative?

Unfortunately not, no.

Are you familiar with the chain rule? Here it is:

$f[g(x)]=f'[g(x)]g'(x)$

This theorem should be memorized, if you haven't already. So, applying it here, we can see the benefits. Our initial function:

$h(u)=\sqrt{9-u^2}$

Let $g(u)=9-u^2$

Then:

$h(u)=\sqrt{g(x)}$

Can you finish it up?
• February 27th 2008, 03:21 PM
mr fantastic
Quote:

Originally Posted by kdogg121
So was I correct then? 3-(1/2)U^-(1/2) is the derivative?

Will this factor?

Sorry, but I can't see how either of the answers you've come up with are the same as the correct answer (see my first reply) of

$h^{'}(u) = -\frac{u}{\sqrt{9 - u^2}}$.

The solution to $h^{'}(u) = 0$ is clearly u = 0, obtained by setting the numerator equal to zero. The factoring business you mentioned is completely irrelevant .......

Therefore:

1. $h^{'}(u) < 0$ when u > 0 AND $-3 < u < 3$ (recall the important note from my earlier reply and also look at where $h^{'}(u) < 0$ is undefined), that is, $0 < u < 3$.

2. $h^{'}(u) > 0$ when yada yada
• February 27th 2008, 03:58 PM
kdogg121
I figured it out with the chain rule finally, thanks guys.
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