#### To determine

**a)To find:**

f is one to one

#### Answer

f is one to one

**Graph of**f:

#### Explanation

Any horizontal line will intersect the given graph at one point only.Therefore, by using the Horizontal Line Test,

**Final statement:** we can conclude that the given function is one to one.

#### To determine

**b)To find:** Derivative at given point

#### Answer

-1/4

#### Explanation

**Calculation:**

**Concept**: By Theorem 7, f-1'a=1f'f-1a, where f'(f-1a)≠0.

First find: The inverse of f.

Write =f(x)⇒y=1x-1 ⇒x=y+1y=1+1y . Thus, f-1x=x+1x=1+1x. Also,f'x= -1x-12.Therefore,f'f-12=f'3/2=-4, which is non-zero.

So, f-1'2=1f'f-12=-14.

#### To determine

**c) To calculate:**

f
−1
(x) and its domain and range for f(x)=1(x−1)

#### Explanation

We have already determined that f-1x=1+1x in Part (b).

**Graph for**f-1:

We will use the fact that if f−1 is the inverse function of f, then the domain of f−1 is the range of f and the range of f−1 is the domain of and all we need to do is to determine the domain and range of f.

**Domain**:

f(x)=1x−1, x>1Domain: (1, ∞)range of f−1: (1, ∞)

#### To determine

**d) To calculate:**

(
f
−1
)'(a) for f(x)=1(x−1)

#### Explanation

**Calculation:**

**Given:**f-1x=x+1x.

Since f-1x=x+1x,f-1'x=-1x2.

So, f-1'2=-14, which is the same value as we calculated in Part (b).

#### To determine

**e) To calculate:**

f and
f
−1
on the same axes for f(x)=1(x−1)

#### Explanation

**Calculation:**

The orange-colored curve represents f while the blue-colored curve represents f-1.