Question

Given that f(x)=x+7 and g(x)=5x+7/x+1, determine when f(x)≥ g(x) using intervals.

Answer 1

f(x)≥ g(x)
x+7 ≥ (5x+7)/(x+1) , I assumed you meant to put those brackets in.
Without the brackets the question would be totally different

clearly g(x) is undefined for x = -1
multiply both sides by x+1

if x+1 > 0, then
x^2 + 8x + 7 ≥ 5x + 7
x^2 + 3x ≥ 0
x(x+3) ≥ 0
we have critical values at x = 0, x = -3 , x = -1

In original : x+7 ≥ (5x+7)/(x+1)
test with a value of x < -3
e.g. let x = -5
-5+7 ≥ (-25+7)/-4
2 ≥ 4.5, false, so x cannot be < -3

test with value between -3 and -1
let x = -2
5 ≥ -3/-1
5 ≥ 3 , Yes, that is true,

test with a value between -1 and 0
let x = -1/2
-1/2 + 7 ≥ (-5/2+7)/(-1/2+1)
6.5 ≥ 9 false, so false between -1 and 0

test for x > 0
x = 5
12 > 32/6 , true, so true for all x > 0

so it looks like we have
-3 ≤ x ≤ -1 OR x > 0

graphing on Demos: www.desmos.com/calculator

confirms my answer is correct

I will let you write this in "interval notation" I am "old school"
and prefer the notation I used.

Answer 2

there are no brackets in the question.