Graph the following showing all working
The main problem i'm having with this one is finding the stationary point and the assymptotes. any help appreciated.
. (the black curve)
1. Domain: . Thus the vertical asymptotes are: . (the blue lines)
2. Horizontal asymptote:
. Thus the horizontal asymtote is: . (the red line)
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:
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 ...)
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
(I used a computer to get this solution)
6. Graph (see attachment): Be careful the axes are scaled differently!