Find algebraically the stationary points and nature of the curve
y=5x^4+4x^3+3x^2+1/2.
How can we do this?
Thanks
2006-10-30 18:15:44 · 4 answers · asked by math 2 in Science & Mathematics ➔ Mathematics
Normally we do this by differenciation. A student has given me this question.I doubt we can do it grafically.
2006-10-30 18:33:22 · update #1
You can't find stationary points by algebra methods only.
You need the derived function to be zero.
y = 5x^4+4x^3+3x^2+1/2.
y ' = 20x^3 + 12x^2 + 6x + 0
x(20x^2 + 12x + 6) = 0
x = 0 or 20x^2 + 12x + 6 = 0
Only at x=0 is a stationary point.
Th
2006-10-30 18:27:04 · answer #1 · answered by Thermo 6 · 0⤊ 0⤋
For the part about the nature of the curve, it has negative gradient when x<0 and positive gradient when x>0. Thus, it is a U-shape like curve, with minimal point at (0, 0.5). No asymptote.
2006-10-31 04:06:10 · answer #2 · answered by back2nature 4 · 0⤊ 0⤋
when x=0 then y=1/2
so (0,1/2) is the stationary point.
2006-10-31 02:37:51 · answer #3 · answered by peterwan1982 2 · 0⤊ 0⤋
dy/dx=20x^3+12x^2+6x
for st pt,dy/dx=0
2x(10x^2+6x+3)=0
x=0 s the only st pt
2006-10-31 02:24:22 · answer #4 · answered by aravind 3 · 0⤊ 0⤋