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 [tex] \frac{1+ \sqrt{5} }{2} [/tex]
e. do ths yourself
try demos graphing calculator
