Dear friends,

The function g is defined by g(x)= , 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.

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- Aug 25th 2010, 06:39 AMstupidguyBonus Question
Dear friends,

The function g is defined by g(x)= , 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 AMProve It
Turning points occur where the derivative is .

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

To differentiate this function, you need to rewrite it as

.

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 AMstupidguy
so

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 AMProve It
I get the same answer.

- Aug 25th 2010, 09:04 AMstupidguy
so we get the turning point which is (1, 1/-16)......how to continue?

- Aug 25th 2010, 02:48 PMmr 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. - Aug 25th 2010, 07:02 PMProve It
- Aug 26th 2010, 04:39 AMstupidguy
@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 AMProve It
Think about .

What happens to when is small? What happens to as gets large?

In this case, . - Aug 26th 2010, 05:36 AMArchie Meade
- Aug 26th 2010, 06:30 AMstupidguyQuote:

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

- Aug 26th 2010, 06:44 AMstupidguygraph
Attachment 18723

attached is the graph....but i need to know how to derive it. - Aug 26th 2010, 06:56 AMArchie Meade
The black parabola is the graph of

The pink asymptotic curve is the graph of

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 AMstupidguy
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 AMProve It