The function ƒ(x) = (x − 1)^2 + 5 is not one-to-one. Find a portion of the domain where the function is one-to-one and find an inverse function. The restricted domain for ƒ is ?

f(x)=(x-1)^2+5 with domain x>1 and range y>5 has inverse g(x)=sqrt(x-5)+1 with domain x>5 and range y>1.

Step-by-step explanation:

The function is a parabola when graphed. It is in vertex form f(x)=a(x-h)^2+k where (h,k) is vertex and a tells us if it's reflected or not or if it's stretched. The thing we need to notice is the vertex because if we cut the graph with a vertical line here the curve will be one to one. So the vertex is (1,5). Let's restrict the domain so x >1.

* if x>1, then x-1>0.

* Also since the parabola opens up, then y>5.

So let's solve y=(x-1)^2+5 for x.

Subtract 5 on both sides:

y-5=(x-1)^2

Take square root of both sides:

Plus/minus sqrt(y-5)=x-1

We want x-1>0:

Sqrt(y-5)=x-1

Add 1 on both sides:

Sqrt(y-5)+1=x

Swap x and y:

Sqrt(x-5)+1=y

x>5

y>1

.

step-by-step explanation:

the answer is d

step-by-step explanation: