Question

PLEASE HELP! 25 PTS!!

Consider function f and g.


`f(x)=(x-16)/(x^2+6x-40)` for
`x\\eq-10` and
`x\\eq 4`


`g(x)=(1)/(x+10)` for
`x\\eq -10`


which expression is equal to f(x) + g(x)?

A.
`(2x-12)/(x^2+6x-40)`

B.
`(2x-20)/(x^2+6x-40)`

C.
`(x-15)/(x^2+6x-40)`

D.
`(x-15)/(x^2+7x-30)`

Answer 1

Solution given:

`f(x)=(x-16)/(x^2+6x-40)`

`g(x)=(1)/(x+10)`

now

f(x)+g(x)=
`(x-16)/(x^2+6x-40)+(1)/(x+10)`....(1)

now

factoring x²+6x-40

we get

x²+10x-4x-40

x(x+10)-4(x+10)

(x+10)(x-4)

now substituting in equation 1 ,we get

f(x)+g(x)=
`(x-16)/((x+10)(x-4))+(1)/(x+10)`

taking l.c.m

=
`((x-16)+(x-4))/((x-10)(x-4))`

=now

opening bracket

`(x-16+x-4)/(x²-10x-4x+40)`

=
`(2x-20)/(x²+6x-40)`

So

answer is :

B.
`\bold b(2x-20)/(x^2+6x-40)`