Help ASAP please! Consider the function f(x)= (x+5)^2 for the domain [-5, infinity]

Question

Consider the function f(x)= (x+5)^2 for the domain [-5, infinity]

Find f^-1(x), where f^-1 is the inverse of f.

Also please state the domain of f^-1 in interval notation!

Thank u!!!

Answer 1

Step 1 : Inverse Function.

f(x) = y = (x+5)^2

Sqrt(y)=x+5

Sqrt(y)-5=x

f^-1(y)=Sqrt(y)-5 then replace y by x

f^-1(x)=Sqrt(x)-5 this is required inverse function.

Step 2: Domain of Inverse Function.

For square root we need non-negative values, so Domain =[0,∞)

Answer 2

f(x) = y = (x+5)^2 with domain [-5, infinity) or x > -5

has vertex (-5,0) it's an upward opening parabola

with range [0, infinity)

the inverse is found by switching x and y

x =(y+5)^2 then solving for the new y

y= sqr(x) -5= f^-1(x)

it's the top half of a horizonal rightward opening parabola with vertex (0,-5) which is also what you get when switching the x and y coordinates of (-5,0),

domain of f^-1(x) is [0,infinity) or x>0

range of f^-1(x) is x> -5 or [-5,0)

just switch the domain and range of f(x) to get the domain and range of f^-1(x)

reflect f(x) over the 45 degree line (x=y) to get f^-1(x), but delete the bottom half of the resulting horizontal parabola to make it a function, otherwise it's an inverse relation, but not an inverse function

while the problem just asked for the "inverse" while not specifying whether its an inverse relation or inverse function, use of "f" in f^-1(x) suggest a function, f for function

the problem as written has the domain of f(x) as [-5,infinity] but that's impossible, it has to be [-5,infinity) with a closed parentheses not bracket, as you can never reach infinity. It's limit but not a real number.