# Bonus Question

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• Aug 25th 2010, 06:39 AM
stupidguy
Bonus Question
Dear friends,

The function g is defined by g(x)= $\frac {1}{(x-5)(x+3)}$, x not equal to 3 and 5.

i) Find the coordinates of the turning point of the curve y=g(x).

ii) Hence, sketch the graph of y=g(x) and write down the range of the function g.
• Aug 25th 2010, 07:10 AM
Prove It
Turning points occur where the derivative is $0$.

So you need to differentiate the function, set it equal to $0$ and solve for $x$. Then back-substitute to find the corresponding $y$.

To differentiate this function, you need to rewrite it as

$g = [(x-5)(x+3)]^{-1}$.

You will now need to use a combination of the Chain Rule and the Product Rule (or expand the inside and just use the Chain Rule).
• Aug 25th 2010, 08:01 AM
stupidguy
so $dy/dx=\frac {2-2x}{(x^2-2x-15)^2}$

$2-2x=0$
$2=2x$
$x=1$
$y=\frac{1}{-16}$

Using Graphmatica, the answer dun match the graph.

Would appreciate if u provide me with the full solution, then I slowly ponder on it. (Wink)
• Aug 25th 2010, 08:11 AM
Prove It
• Aug 25th 2010, 09:04 AM
stupidguy
so we get the turning point which is (1, 1/-16)......how to continue?
• Aug 25th 2010, 02:48 PM
mr fantastic
Quote:

Originally Posted by stupidguy
Dear friends,

I am too poor to afford maths tuition. Please teach me how to do this simple sum.

The function g is defined by g(x)= $\frac {1}{(x-5)(x+3)}$, x not equal to 3 and 5.

i) Find the coordinates of the turning point of the curve y=g(x).

ii) Hence, sketch the graph of y=g(x) and write down the range of the function g.

*Ahem* Attention all posters (including the original poster):

Since this question is posted in the PRE-calculus subforum, the expectation is that the question is to be done without using calculus.

Draw the graph of the quadratic function y = (x - 5)(x + 3). Now think about what happens to this graph when you take the reciprocal of the y-coordinates for each x-coordinate.
• Aug 25th 2010, 07:02 PM
Prove It
Quote:

Originally Posted by mr fantastic
*Ahem* Attention all posters (including the original poster):

Since this question is posted in the PRE-calculus subforum, the expectation is that the question is to be done without using calculus.

Draw the graph of the quadratic function y = (x - 5)(x + 3). Now think about what happens to this graph when you take the reciprocal of the y-coordinates for each x-coordinate.

And yet, when given the advice to use the derivative, the OP knew exactly what to do...

This is why I say there should be a link to the Calculus forum in the High School section of this forum...
• Aug 26th 2010, 04:39 AM
stupidguy
@Prove It: good suggestion.

@ mrfantastic sry, I do not know the difference between calculus and pre-calculus. However, I I just want someone to teach me how to tackle this question. I dun get it ur "advice"
• Aug 26th 2010, 05:29 AM
Prove It
Think about $\frac{1}{n}$.

What happens to $\frac{1}{n}$ when $n$ is small? What happens to $\frac{1}{n}$ as $n$ gets large?

In this case, $n = (x - 5)(x + 3)$.
• Aug 26th 2010, 05:36 AM
Quote:

Originally Posted by stupidguy
@Prove It: good suggestion.

@ mrfantastic sry, I do not know the difference between calculus and pre-calculus. However, I I just want someone to teach me how to tackle this question. I dun get it ur "advice"

$(x-5)(x+3)$

has zeros at $x=5$

and $x=-3$

It's a U-shaped graph and the minimum occurs halfway between the zeros.
If you invert the graph to

$\displaystyle\frac{1}{(x-5)(x+3)}$

we now have vertical asymptotes at $x=5$ and $x=-3$

but the turning point still occurs at the same x we previously had the minimum.
This is because..

$\displaystyle\frac{1}{most\ negative}=least\ negative$

or

$\displaystyle\frac{1}{least\ positive}=most\ positive$
• Aug 26th 2010, 06:30 AM
stupidguy
Quote:

If you invert the graph to

we now have vertical asymptotes at and

but the turning point still occurs at the same x we previously had the minimum.
This is because..

or

still dun get the link. have not clearly answered my initial question.
• Aug 26th 2010, 06:44 AM
stupidguy
graph
Attachment 18723

attached is the graph....but i need to know how to derive it.
• Aug 26th 2010, 06:56 AM
Quote:

Originally Posted by stupidguy
Attachment 18723

attached is the graph....but i need to know how to derive it.

The black parabola is the graph of

$(x-5)(x+3)$

The pink asymptotic curve is the graph of

$\displaystyle\frac{1}{(x-5)(x+3)}$

The y-axis is not suitably scaled to effectively view the pink curve.
If we "stretch" it, you'd more easily see that the local maximum (turning point) of the pink graph
occurs at the exact same x which locates the minimum value (turning point) of the U-shaped parabola.

If you invert a quadratic equation, the zeros go towards infinity (become vertical asymptotes),
local maxima become local minima and local minima become local maxima.
• Aug 26th 2010, 07:11 AM
stupidguy
not clear. I can only understand that x=5 and x=-3 are vertical asymptotes, (1,-16) is the max point without the use of the U-shaped parabola.

Someone end this misery by giving me the complete steps and answers to my question. Dun just hint!
• Aug 26th 2010, 07:43 AM
Prove It
Quote:

Originally Posted by stupidguy
not clear. I can only understand that x=5 and x=-3 are vertical asymptotes, (1,-16) is the max point without the use of the U-shaped parabola.

Someone end this misery by giving me the complete steps and answers to my question. Dun just hint!

We are not a homework service. You need to make an effort to understand this material yourself.
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