# Thread: calculus graph, please check, thanx

1. ## calculus graph, please check, thanx

Qu.

Use graph sketching strategy to sketch the graph of the function f(x) = 4x -15/ x^2 -9

Step 1 The domain of f is all natural numbers except + 3 or - 3 when x^2 - 9 = 0

Step 2 Function f in nether even or odd

step 3 x intercept , 4x -15 = 0, x = 3 3/4
y intercept f (0) = 5/3

step 4

f(x) = 4x -15/ (x-3) (x+3)

f is positive (3 3/4, infinite), and negative on (-infinite, -3), (-3,3), and (3, 3 3/4)

Step 5

the derivative of f(x) = -4x^2 -36 +30x/ (x^2-9)^2

2. Are you wanting the derivative of:

$\frac{-4x^{2}+30x-36}{(x^{2}-9)^{2}}$?.

Rewrite as such and use the product rule:

$(-4x^{2}+30x-36)(x^{2}-9)^{-2}$

$(-4x^{2}+30x-36)(-2)(x^{2}-9)^{-3}(2x)+(x^{2}-9)^{-2}(-8x+30)$

$\frac{16x^{3}-120x^{2}+144x}{(x^{2}-9)^{3}}+\frac{-8x+30}{(x^{2}-9)^{2}}$

Multiply the top and bottom of the right side by $(x^{2}-9)$

and simplify:

$\Large\frac{2(4x^{3}-45x^{2}+108x-135)}{(x^{2}-9)^{3}}$

3. Originally Posted by fair_lady0072002
Qu.

Use graph sketching strategy to sketch the graph of the function f(x) = 4x -15/ x^2 -9

Step 1 The domain of f is all natural numbers except + 3 or - 3 when x^2 - 9 = 0

Step 2 Function f in nether even or odd

step 3 x intercept , 4x -15 = 0, x = 3 3/4
y intercept f (0) = 5/3

step 4

f(x) = 4x -15/ (x-3) (x+3)

f is positive (3 3/4, infinite), and negative on (-infinite, -3), (-3,3), and (3, 3 3/4)

Step 5

the derivative of f(x) = -4x^2 -36 +30x/ (x^2-9)^2

Sketching curves has always been one of my favorite thing in calculus. I will you show you my way.
-------------------------------------------
Step 1: Find domain
Find the domain of the function,
$f(x)=\frac{4x-15}{x^2-9}$
Note, $x^2-9\not = 0$ thus, $x\not = \pm 3$

Step 2: Symettry
Does $f(-x)=f(x)$
You can see that,
$\frac{4(-x)-15}{(-x)^2-9}=\frac{-4x-15}{x^2-9}$--->No
Does $f(-x)=-f(x)$
You can see that,
$\frac{4(-x)-15}{x^2-9}=\frac{-4x-15}{x^2-9}$ and that,
$-f(x)=\frac{-4x+15}{x^2-9}$---> No

Step 3: Asymptotes
Yes it does at $x=-3,3$ (since numerator is non-zero while denominator is zero).

I countinue later, it takes longgg time.

4. ## ThePerfectHacker- thanks, continued

Horizontal asymptote

y = 0 ?

When m < n, the horizontal asymptote is the line y = 0

where m is the degree of polynomial in the numerator and n is the degree of polynomial in the denominator.

5. Originally Posted by ThePerfectHacker
Sketching curves has always been one of my favorite thing in calculus. I will you show you my way.
-------------------------------------------
Step 1: Find domain
Find the domain of the function,
$f(x)=\frac{4x-15}{x^2-9}$
Note, $x^2-9\not = 0$ thus, $x\not = \pm 3$

Step 2: Symettry
Does $f(-x)=f(x)$
You can see that,
$\frac{4(-x)-15}{(-x)^2-9}=\frac{-4x-15}{x^2-9}$--->No
Does $f(-x)=-f(x)$
You can see that,
$\frac{4(-x)-15}{x^2-9}=\frac{-4x-15}{x^2-9}$ and that,
$-f(x)=\frac{-4x+15}{x^2-9}$---> No

Step 3: Asymptotes
Yes it does at $x=-3,3$ (since numerator is non-zero while denominator is zero).

I countinue later, it takes longgg time.

6. Originally Posted by fair_lady0072002
With what?
Tell me extacly you were able to do (post it I will check it) and then procede to tell me what you do not know how to solve.

7. ## ThePerfectHacker - Graph of a function

Found derivative of f

= (-4x +6)(x-6)/(x^2-9)^2

Problems establishing when f is increasing or decreasing and if any stationary points

8. Originally Posted by fair_lady0072002
Found derivative of f

= (-4x +6)(x-6)/(x^2-9)^2

Problems establishing when f is increasing or decreasing and if any stationary points
Fair Lady Check out the thread below

9. Originally Posted by fair_lady0072002
Found derivative of f

= (-4x +6)(x-6)/(x^2-9)^2

Problems establishing when f is increasing or decreasing and if any stationary points
Multiply the numerator (open parantheses),
$\frac{-4x^2+30x-36}{(x^2-9)^2}$
Use quotient rule,
$\left( \frac{u}{v} \right) = \frac{u'v-uv'}{v^2}$
Thus,
$\frac{(-4x^2+30x-36)'[(x^2-9)^2]-(-4x^2+30x-36)[(x^2-9)^2]'}{(x^2-9)^4}$ (1)
Note that,
$(-4x^2+30x-36)'=-8x+30$
And,
$[(x^2-9)^2]'=2(2x)(x^2-9)=4x(x^2-9)$--->Rule of chains.

Now rewrite expression (1) without derivatives,
$\frac{(-8x+30)(x^2-9)^2-4x(-4x^2+30x-36)(x^2-9)}{(x^2-9)^4}$
Factor, that the term $(x^2-9)$ thus,
$\frac{(x^2-9)[(-8x+30)(x^2-9)-4x(-4x^2+30x-36)]}{(x^2-9)^4}$
Cancel,
$\frac{(-8x+30)(x^2-9)-4x(-4x^2+30x-36)}{(x^2-9)^3}$
Open ye olde' parantheses,
$\frac{-8x^3+30x^2+72x-270+16x^3-120x^2+144x}{(x^2-9)^3}$
Thus, combine,
$\frac{8x^3-90x^2+192x-270}{(x^2-9)^3}$

10. Originally Posted by ThePerfectHacker
...
Open ye olde' parantheses,
$\frac{-8x^3+30x^2+72x-270+16x^3-120x^2+144x}{(x^2-9)^3}$
Thus, combine,
$\frac{8x^3-90x^2+192x-270}{(x^2-9)^3}$

$\frac{8x^3-90x^2+216x-270}{(x^2-9)^3}$ as galactus has calculated already