Its urgent (45 ! )

suppose f(x)=√x+1

a. what are the domain and range of f?
b. find f^-1(x). what are its domain and range?
c. show that (fof^-1)(a)=a=(f^-1of)(a) for any a in the respective domains
d. solve the equation f(x)=f^-1(x). remember to check for extraneous roots
e. graph the functions f and f^-1. be sure that you accurately represent the domains of each function. interpret graphically the solutions you found to the equation in part (d).

A.
domain is th enumbers you can use
remember you cannot take square root of a negative
also you cannot divide by 0

domain is all real numbers except those that result in a square root of a negative or division by 0
domain is how far left or right the graph goes

range is the output from inputing the domain, it is how far up and down the graph goes

a. domain is
hmm, squaerr root
see under parenthasees
x+1
so x≥0 is the domain
domain is x≥0
so then the range is from 0 to infinity or y≥0

b. replace f(x) with y and switch x and y and solve for y
y=√(x+1)
x=√(y+1)
square
x²=y+1
x²-1=y
f⁻¹(x)=x²-1
domain is all real numbers
range, this is a parabola, when it gets to x=0, y=-1 that is minimum
range is y≥-1 or from -1 to infinity

basicaly show that f(f⁻¹(x))=x=f⁻¹(f(x))

so if you did that
f(f⁻¹(x))=√(x²-1+1)
f(f⁻¹(x))=√(x²)
f(f⁻¹(x))=x

reverse

f⁻¹(f(x))=(√(x+1))²-1
f⁻¹(f(x))=x+1-1
f⁻¹(f(x))=x

d.
√(x+1)=x²-1
squaer both sides
x+1=x⁴-2x²+1
minus x both sides and minus 1
0=x⁴-2x²-x
factor
0=(x)(x+1)(x²-x-1)
x=0 and x=-1 and 0.5+0.5√5 and 0.5-0.5√5
test them
wait, the last one isn't in the domain
if we did x=0, then we get 1=-1, false

the soltions are x=-1 and

e. do ths yourself
try demos graphing calculator

Rounding decimals is very similar to rounding other numbers. if the hundredths and thousandths places of a decimal is forty-nine or less, they are dropped and the tenths place does not change. for example, rounding 0.843 to the nearest tenth would give 0.8. , your welcome !