Who can help me with this maths problem?
2006-12-17 06:36:47 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
y=2/(x-1)-2/(x-1)^2
= ( 2(x-1) -2 ) /(x-1)^2
=( 2x -2-2) /(x-1)^2
= (2x-4) /(x-1)^2
y'= ( (x-1)^2 (2) - (2x-4) 2(x-1) )/(x-1)^4
= (x-1) ( 2x-2 -4x+ 8) /(x-1)^4
= ( -2x + 6) /(x-1)^3
y'=0 if -2x + 6 =0 so x=3, which is the only critical point.
also notice that y and y' are not defined for x=1.
.
2006-12-18 13:41:50 · answer #1 · answered by Anonymous · 4⤊ 0⤋
y=2/(x-1)-2/(x-1)^2 adding them we get:
= ( 2(x-1) -2 ) /(x-1)^2
=( 2x -2-2) /(x-1)^2
= (2x-4) /(x-1)^2
y'= ( (x-1)^2 (2) - (2x-4) 2(x-1) )/(x-1)^4
= (x-1) ( 2x-2 -4x+ 8) /(x-1)^4
= ( -2x + 6) /(x-1)^3
y'=0 if the numerator -2x + 6 =0 => x=3 is the only critical point.
2006-12-19 17:50:11 · answer #2 · answered by lobis3 5 · 3⤊ 0⤋
y=2/(x-1)-2/(x-1)^2 = ( 2(x-1) -2 ) /(x-1)^2
=( 2x -2-2) /(x-1)^2 = (2x-4) /(x-1)^2
y'= ( (x-1)^2 (2) - (2x-4) 2(x-1) )/(x-1)^4
= (x-1) ( 2x-2 -4x+ 8) /(x-1)^4 = ( -2x + 6) /(x-1)^3
y'=0 <=> -2x + 6 =0 <=> x=3
2006-12-20 11:04:13 · answer #3 · answered by pedazodegente 1 · 2⤊ 0⤋
the expression
=[2(x-1)-2]/(x-1)^2
=2x-2-2/(x-1)^2
=2(x-2)/(x-1)^2
f'(x)=2[(x-1)^2-(x-2)*2(x-1)]/(x-1)^4
setting this to zero
2[x^2-2x+1-2x^2-6x+4]=0
-x^2-8x+5=0
=>x^2+8x-5=0
use the Quadratic formula to find the points
2006-12-17 14:45:27 · answer #4 · answered by raj 7 · 1⤊ 0⤋