Math.i don't know how to do, please teach me and show step.?

Question

if g(x)= x^2-1 then g(x-1)-g(x+1)=?

Answer 1

g( x - 1) = ( x - 1 )^2 - 1 = x^2 - 2x + 1 - 1 = x^2 - 2x

g( x + 1) = ( x + 1 )^2 - 1 = x^2 + 2x + 1 - 1 = x^2 + 2x

g ( x - 1 ) - g ( x + 1) =
[ x^2 - 2x ] - [ x^2 + 2x ] = - 4x

Answer 2

if
g(x) = x^2-1

first, find g(x - 1)
g(x - 1) = (x - 1)^2 - 1
= (x - 1)(x - 1) -1
= x^2 - x - x + 1 - 1
= x^2 - 2x

then, find g(x + 1)
g(x + 1) = (x + 1)^2 - 1
= (x + 1)(x + 1) - 1
= x^2 + x + x + 1 - 1
= x^2 + 2x

finally....
g(x - 1) - g(x + 1) = x^2 - 2x - x^2 + 2x
= x^2 - x^2 - 2x + 2x
= 0

therefore the answer is 0...

Answer 3

g(x) = x^2 - 1

g(anything) = (anything)^2 - 1

g(x - 1) = (x - 1)^2 - 1
g(x + 1) = (x + 1)^2 -1

g(x-1) -g(x+1) = (x - 1)^2 - 1 - (x + 1)^2 + 1 = (x - 1)^2 - (x + 1)^2

This can be further simplified by expanding the binomials

Answer 4

Others have helped you with the algebra. I would just like to point out to Mis Hottie that someone who doesn't know the difference between "your" and "you're" should be a little careful when accusing others of being "dumb" ...

Answer 5

just substitute "g" with "x^2-1" in the equation. from there its just algebra

Answer 6

if g(x)=(x^2)-1
then
g(x-1)-g(x+1)= [{(x-1)^2}-1] - [{(x+1)^2}-1]
={(x-1-1)(x-1+1)}-{(x+1-1)(x+1+1)}
={x(x-2)}-{x(x+2)}
=[(x^2)-2x]-[(x^2)+2x]
=(-4x)


using formula (a^2-b^2)=(a+b)(a-b)

Answer 7

g(x)=x^2-1
g(x-1)=(x-1)^2-1 now (a-b)^2=a^2-2 ab+b^2
=(x^2-2x+1)-1 now 1-1=0
g(x-1)=(x^2-2x)
g(x+1)=(x+2)^-1 now (a+b)^2=a^2+2 ab+b^2
=(x^2+2x+1)-1 now 1-1=0
g(x+1)=x^2+2x
g(x-1)-g(x+1)=(x^2-2x)-(x^2+2x)
=x^2 - 2x - x^2 - 2x now-2x+2x=0
=-2x-2x
=-4x
-4x

Answer 8

g(x)= x^2-1
g(x-1) = (x-1)^2-1
g(x-1) = (x-1)(x-1)-1
g(x-1)=x^2-2x+1-1
g(x-1)=x^2-2x
g(x+1)=(x+1)^2-1
g(x+1)=(x+1)(x+1)-1
g(x+1)=x^2+2x+1-1
g(x+1)=x^2+2x
[x^2-2x] - [x^2+2x] = -4x

Answer 9

i ABSOLUTELY agree with mis hottie's comment.