For F(x) = 3x^3 - 1/x^2
a) find F'(x), F''(x), F'''(x), F''''(x)
b) will the 20th derivative have a positive integer or negative coefficient?
2007-05-06 08:40:54 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Since F(x) = 3x^3 - 1/x^2
So F(x) = 3x^3 - x^(-2)
F'(x) = 9x^2 +2x^(-3)
F''(x) = 18x - 6x^(-4)
F'''(x) = 18 +24x^(-5)
F''''(x) = -120 x^(-6)
After that even dreivatives have a -ve coeffecient and odd ones have a +ve coeffecient
So the 20th derivative have a negative coefficient
2007-05-06 08:53:24 · answer #1 · answered by a_ebnlhaitham 6 · 0⤊ 0⤋
F´=9x^2+2/x^3
F´´= 18x -(2*3)/x^4
F´´´=18 +(4!)/x^5
F´´´´=-(5!)/x^6
F(n)(x)=(-1)^n+1 * (n+1)! *1/x^n+2 for n>=4 so if n= 20 the coefficient would be negative
2007-05-06 08:57:05 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋
The by-product of any consistent is 0 so both will change into 0. ur equation is now 0 - by-product of(2/x) u can rewrite -2/x as -2x^-a million convey the -a million down and multiply it through -2. also subtract the exponent through 1 million ur answer: 2x^-2
2016-12-05 10:53:37 · answer #3 · answered by allateef 4 · 0⤊ 0⤋
1. f'.... 9x^2 + 2/x^3
2. f'' ..... 18x - 6/ x^4
3. f'''..... 18 + 24/x^5
4. f''''..... 0 - 120/x^6
from then all even is negative and odd is positive..
then... b) the 20th is negative
2007-05-06 08:50:46 · answer #4 · answered by Anonymous · 0⤊ 0⤋
the 3x^3 term is unimportant what is more so is the x^-2 term
for the 1st derivative it will be +ve, 2nd -ve, 3rd +ve and so on so the 20th will be -ve
2007-05-06 08:51:13 · answer #5 · answered by SS4 7 · 0⤊ 0⤋
Keep in mind that everytime you take the derivative of this polynomial function, the degree of the polynomial will reduce by one - so the fourth derivative will be a constant. After that, your derivatives are zero -
Hope this gives you a hint -
2007-05-06 08:48:51 · answer #6 · answered by seadreamer164 2 · 0⤊ 2⤋