ALGEBRA: CAN YOU SOLVE?

(f+g)(x)=4(root x) +3(root x-1) This is in simplest form.

find ftg? do you mean (f)times(g)? If so,then its:

12 (root x*(x-1))

I'm not so sure about the last one but I think

(f)/(g)=(4(root x(x-1))) over (3(x-1))

Given I assume

f(x) = 4√ ̅x ..... quad root

g(x) = 3√ ̅(x-1) ... cube root

find (f*g)(x), (f/g)(x), (f+g)(x)

simplify answer

(1)

(f*g)(x)

= (4√ ̅x) * [3√ ̅(x-1)]

= x^(1/4) * (x-1)^(1/3)

= x^(3/4*3) * (x-1)^(4/3*4)

= x^(3/12) * (x-1)^(4/12)

= [x^3 * (x-1)^4]^(1/12)

.... 1

... 1 1

.. 1 2 1

. 1 3 3 1

1 4 6 4 1

(f*g)(x)

= [x^3*(x^4 - 4x^3 + 6x^2 - 4x + 1)]^(1/12)

= (x^7 - 4x^6 + 6x^5 - 4x^4 + x^3)^(1/12)

(2)

(f/g)(x)

= (4√ ̅x) / [3√ ̅(x-1)]

= x^(1/4) / (x-1)^(1/3)

= [x^3 / (x-1)^4]^(1/12)

= {x^4 / [x(x-1)^4}^(1/12)

= {(1/x) * [x/(x-1)]^4}^(1/12)

= x^(-1/12) * [x/(x-1)]^(4/12)

(3)

(f+g)(x)

= 4√ ̅x + 3√ ̅(x-1)

= x^(1/4) + (x-1)^(1/3)