f(x)=x^4-4x^3+2
2007-10-21 03:01:10 · 4 answers · asked by nala 1 in Science & Mathematics ➔ Mathematics
f´(x) = 4x^3-12x^2 = 4x^2(x-3)=0 x= 0 and x=3
f´´(x) = 12x^2-24x At x= 0 f´´=0 not applicabe but f´does not change sign at x=0 so there no max nor min
f¨¨(3) =84-72 >0 local min.
2007-10-21 03:09:55 · answer #1 · answered by santmann2002 7 · 0⤊ 0⤋
f'(x) = 4x^3 - 12x^2
....... = 4x^2(x - 3)
which is zero when
x = 0 or 3.
We know x = 0 gives a point of inflexion since it's a double root, but check anyway:
f"(x) = 12x^2 - 24x
...... = 12x(x - 2)
which changes sign as x goes from small negative to small positive, hence (0, 2) is a stationary inflexion.
f"(3) = 36. Since this is positive,
(3, -25) is a minimum turning point. This is the only relative extremum.
2007-10-21 10:12:01 · answer #2 · answered by Hy 7 · 0⤊ 0⤋
I don't really feel like trying to figure out the question since I have my own to do, but one thing I can tell you is that there is in fact at least a minimum. Just plug the equation into a graphing calculator and you'll see it.
2007-10-22 19:24:37 · answer #3 · answered by Calc student 1 · 0⤊ 0⤋
4x^3-3x^2=0
12x^2-6x=0
x(12x-6)=0
therfore, x may be 0 or .5
2007-10-21 10:10:27 · answer #4 · answered by sathyanarayanan k 2 · 0⤊ 0⤋