If f (x) = 2x^2+1 and g (x) = x^2-7 find (f+g) (x)

Answer:f=-

7



x

2

+ 1-2g-x



Step-by-step explanation:

1 Subtract {x}^{2}x

2

from both sides.

2{x}^{2}+1-g-x-{x}^{2}=-7f+g2x

2

+ 1-g-x-x

2

= -7f+g

2 Simplify 2{x}^{2}+1-g-x-{x}^{2}2x

2

+ 1-g-x-x

2

to {x}^{2}+1-g-xx

2

+ 1-g-x.

{x}^{2}+1-g-x=-7f+gx

2

+ 1-g-x=-7f+g

3 Subtract gg from both sides.

{x}^{2}+1-g-x-g=-7fx

2

+ 1-g-x-g=-7f

4 Simplify {x}^{2}+1-g-x-gx

2

+ 1-g-x-g to {x}^{2}+1-2g-xx

2

+ 1-2g-x.

{x}^{2}+1-2g-x=-7fx

2

+ 1-2g-x=-7f

5 Divide both sides by - 7-7.

-/frac{{x}^{2}+1-2g-x}{7}=f-

7



x

2

+ 1-2g-x

= f

6 Switch sides.

f=-/frac{{x}^{2}+1-2g-x}{7}f=-

7



x

2

+ 1-2g-x

F (x) + g (x) = [2x^2 + 1] + [x^2 - 7]

≈ 3x^2 - 6

Hence (f + g) (x) = 3x^2 - 6