Graph the following showing all working

4X^3/((X-4)^2(X+2))

The main problem i'm having with this one is finding the stationary point and the assymptotes. any help appreciated.

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- Aug 13th 2007, 11:34 PMmathsBHere's another (harder) one
Graph the following showing all working

4X^3/((X-4)^2(X+2))

The main problem i'm having with this one is finding the stationary point and the assymptotes. any help appreciated. - Aug 14th 2007, 05:14 AMearboth
Hello,

$\displaystyle f(x)=\frac{4x^3}{(x-4)^2 \cdot (x+2)}$. (the**black**curve)

1. Domain: $\displaystyle D = \mathbb{R} \setminus \{-2, 4\}$. Thus the vertical asymptotes are: $\displaystyle \boxed{x = 4~ or~ x = -2}$ . (the blue lines)

2. Horizontal asymptote:

$\displaystyle \lim_{|x| \rightarrow \infty}f(x) = \lim_{|x| \rightarrow \infty}\left(\frac{4x^3}{(x-4)^2 \cdot (x+2)} \right) = \lim_{|x| \rightarrow \infty}\left(\frac{4x^3}{x^3-6x^2+32}\right) = 4$. Thus the horizontal asymtote is: $\displaystyle \boxed{y = 4}$ . (the red line)

3. Zeros:

f(x) = 0 if 4x³ = 0 that means there is one zero at x = 0.

4. Stationary points. Calculate the 1st drivative of f. Use quotient rule:

$\displaystyle f'(x) = \frac{(x-4)^2 \cdot (x+2) \cdot 12x^2 - 4x^3\cdot((x-4)^2+(x+2)\cdot 2 \cdot (x-4))}{\left( (x-4)^2 \cdot (x+2)\right)^2}$ = $\displaystyle \frac{-24x^2(x+4)}{(x-4)^3 \cdot (x+2)^2}$

f'(x) = 0 if the numerator equals zero. Thus you get 2 stationary points at x = 0 or x = -4. Minimum at x = -4: f(-4) = 2

5. Points of inflection:

Calculate the 2nd drivative of f. (You'll need at least half an hour :D ...)

$\displaystyle f''(x) = \frac{48x(x^3+8x^2+16x+32)}{(x-4)^4 \cdot (x+2)^3}$

f''(x) = 0 if the numerator equals zero. First you get x = 0 that means at x = 0 is**no**stationary point but a point of inflection with a horizontal slope. (Such a point is called in German a terrace point).

The 2nd solution you'll get if

$\displaystyle x^3+8x^2+16x+32 = 0 ~\Longrightarrow~x\approx -6.26079...$ (I used a computer to get this solution)

6. Graph (see attachment): Be careful the axes are scaled differently! - Aug 14th 2007, 09:50 PMmathsB
Thanx, Gr8 help :):)