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 [tex] \frac{1+ \sqrt{5} }{2} [/tex]
