f(x) = √(x+1) - √x

Question

Find the domain of the following

Answer 1

one method is plug the function into a graphiing calculator and look at the x values

it's a curve with minimum x coordinate = 0

the domain is x> 0

f(x) = sqr(x+1) - sqr(x)

x+1 cannot be negative. x+1>1, x>-1 and

x cannot be negative, x>0

net result is x>0

domain is x>0

x is in the interval [0, infinity)

this assumes you didn't mean f(x) = sqr[(x+1)-sqr(x)]. That seems most unlikely

but if you somehow did mean that, then

again x>0 due to the 2nd square root term

but also x+1 >sqr(x)

that is always true for x= or >0

so the domain is still the same, x>0

square both sides

(x+1)^2 > [sqr(x)]^2

x^2 +2x +1 > x

x^2 +x +1 > 0, discriminant <0, so no real solutions

x> -1/2 +/- (1/2)sqr(1-4) but those are imaginary numbers

this is the solution to when x+1 = sqr(x), which is never when the domain is restricted to real numbers

graph the function, and x is always non-negative, and the graph is always in Quadrant I