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