# What is x if f(x)=2x^2-1 and g(x)=x-1 if fog=0

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have f(x)=2x^2-1 and g(x)= x-1,

fog(x )= f(g(x))

= f( x-1)

= 2(x-1)^2 - 1

=> 2*(x^2 + 1 - 2x) - 1

=> 2x^2 + 2 - 4x - 1

=> 2x^2 - 4x + 1

As fog(x ) = 0

=> 2x^2 - 4x + 1 = 0

x1 = [ -b + sqrt (b^2 - 4ac)]/2a

=> x1 = [ 4 + sqrt (16 - 8)]/4

=> x1 = 1 + sqrt 8/4

x2 = 1 - sqrt 8 / 4

So x = 1 + sqrt 8/4 and x = 1 - sqrt 8 / 4

neela | High School Teacher | (Level 3) Valedictorian

Posted on

What is x if f(x)=2x^2-1 and g(x)=x-1 if fog=0.

f(x) = 2x^2 - 1.

fog(x) = 2(g(x))^2 - 1.

fog(x) = 2(x-1)^2 - 1.

fog(x) = 2(x^2-2x+1) - 1.

Therefore fog(x) = 0 ==>  2(x-1)^2 - 1 =0.

==>  2(x-1)^2 = 1.

==> (x-1)^2 = 1/2.

==> x=1 = (1/2)^(1/2), or (x-1 = -(1/2)^(1/2).

==>x = 1+(1/2)^(1/2) , Or x= 1-(1/2)^(1/2).

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To determine x, we'll have to find (fog)(x).

(fog)(x) = f(g(x))

We'll substitute x by g(x):

f(g(x)) = 2[g(x)]^2 - 1

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

We'll expand the square:

f(g(x)) = 2x^2 - 4x + 2 - 1

We'll combine like terms:

f(g(x)) = 2x^2 - 4x + 1

We'll put f(g(x)) = 0:

2x^2 - 4x + 1 = 0