# Calculus Graph Sketching strategy.

• May 27th 2014, 06:06 AM
michele
Calculus Graph Sketching strategy.
Hello all,

I'm new here and I taking a course in calculus what is totally new to me, I having been work trough differentiation and now we got to do some graph sketching following 6 steps described in the book I have started this one but I'm not sure if I'm doing it correctly so all explanation and support in how to tackle and understand it is very much appreciated. So here we go

The Function f(x)= 6x-5 / 49x2

STEP 1 - Identify the domain of f

I came up with the domain is all R once 49x2 is never 0.

Is it correct?

STEP 2 - Wether f is even or odd

following my examples:

f(-x) = 6(-x)-5 / 4-9(-x)2
= -6x-5 / 4-9(x)2

Thus -f(x) the function is odd

STEP 3 - x and y intercepts

The x intercept is 6x-5=0
6x=5
x=5/6
x = 0.833
The y intercept is f(0) = 6(0)-5 / 4-9(0)2
= -5 / 4
= -1.25

STEP 4 - The Intervals where x is positive or negative

f(x) = 6x-5 / 4-9x2
= 6x-5 / (2-3x)(2+3x)

 x (-∞,-1.25) -1.25 (-1.25,0.833) 0.833 (0.833,∞) 6x-5 - - - - + 2-3x + - - - - 2+3x - + + + + f(x) + + + + -

Thus f is positive on the Interval (-∞,-1.25) and negative on the interval(0.833,∞).

STEP 5
- The Intervals where f is increasing or decreasing and any stationary points, local maxima and local minima.

f ' (x) = (6)(-9x2+4) - (-18x)(6x-5) / (4-9x2)2

= 54x-66 / (4-9x2)2

 x (-∞,-1.25) -1.25 (-1.25,0.833) 0.833 (0.833,∞) 54x2-66 - + - - + (4-9x2)2 + + + + + f ' (x) - + - - +

Thus f is decreasing on
(-∞,-1.25)
and increasing on
(0.833,∞)
also f has a stationary point at
0.833
which is a local minim with value f(0.833) = 0 and a stationary point at -1.25 which is a local maximum with value f(-1.25) = 1.242

STEP 6 - The asymptotic behaviour of f.

I can't understand how to discover the asymptote behaviours.

Well this is it does it all make sense to anyone? If it does what I have done is correct? If not please I really need to understand it and a step by step will really be appreciated.

Thanks.
• Jun 3rd 2014, 07:08 AM
michele
Re: Calculus Graph Sketching strategy.
Please know one can give me some advices so I know if it is wrong or right?
• Jun 3rd 2014, 08:05 AM
Plato
Re: Calculus Graph Sketching strategy.
Quote:

Originally Posted by michele
Please know one can give me some advices so I know if it is wrong or right?

Frankly. I for one find your notation so difficult to read. So I just did not.

In fact, I not even sure what the function is. Is it $f(x)=\dfrac{6x-5}{4-9x^2}~?$

Clearly that in a lot easier to read. Learning to use LaTeX.

The code \$f(x)=\dfrac{6x-5}{4-9x^2}\$ gives $f(x)=\dfrac{6x-5}{4-9x^2}$
• Jun 7th 2014, 06:58 AM
guinster
Re: Calculus Graph Sketching strategy.
Hi
I think you may have missed the asymptotes on the y axis where f is undefined.
These occur where the denominator tends to zero.
If you solve 4-9x^2 = 0 it gives +/- 2/3.
So domain is R - {-2/3 , 2/3} and graph approaches infinity either side of these two points like 1/x does at x = 0.
• Jun 25th 2014, 11:07 AM
guinster
Re: Calculus Graph Sketching strategy.
Hi
If you draw two vertical lines on your graph - one at y = -2/3 and the other at y = 2/3 these are the asymptotes.
The curve will approach either infinity or - infinity either side of the asymptote.

I have done some sketches.

Hope they help.

Attachment 31181

Regards
• Jun 25th 2014, 03:23 PM
michele
Re: Calculus Graph Sketching strategy.
My sketch looks similar to yours :)

But when I checked on my pc the graph should look like this:
Is it correct?
• Jun 25th 2014, 04:48 PM
topsquark
Re: Calculus Graph Sketching strategy.
I'm coming in late to the party. Step 2...$\displaystyle \frac{6x - 5}{4 - 9x^2}$ is neither even nor odd, as your graph shows.

-Dan
• Jun 25th 2014, 10:38 PM
michele
Re: Calculus Graph Sketching strategy.
An how can I prove it is neither even or odd?
• Jun 26th 2014, 09:21 AM
topsquark
Re: Calculus Graph Sketching strategy.
The same way you did. You just didn't come to the right conclusion.

A function is even if f(-x) = f(x). A function is odd if f(-x) = -f(x). In your case:
$\displaystyle f(x) = \frac{6x - 5}{4 - 9 x^2}$ implies that

$\displaystyle f(-x) = \frac{6(-x) - 5}{4 - 9 (-x)^2}$

$\displaystyle f(-x) = \frac{-6x - 5}{4 - 9 x^2} \neq f(x) \text{ nor } -f(x)$

-Dan
• Jun 26th 2014, 09:51 AM
michele
Re: Calculus Graph Sketching strategy.
Thanks topsquark, so with this conclusion my graph now is valid?
• Jun 26th 2014, 11:46 AM
guinster
Re: Calculus Graph Sketching strategy.
Hi

My sketches (A) to (D) where meant to cover the four possible types of asymptote.

Your graph is showing that the behaviour at x = -2/3 is as in my sketch (D) and
the behaviour at x = 2/3 is as in my sketch (C).

Regards
• Jun 26th 2014, 11:48 PM
michele
Re: Calculus Graph Sketching strategy.
Thanks Again Guisnster.

Now I fell more confident about my work.