#### To determine

**a)To find:**

f is one to one.

#### Answer

f is one to one.

#### Explanation

The graph of f looks like:

You can determine whether or not the given function is one to one by just looking at the graph with the help of Horizontal Line Test. Horizontal Line Test says that a function f is one to one if and only if any horizontal line intersects the graph of given function exactly at one point. In other words, the Horizontal Line Test can be stated as:

A function f is NOT one to one if and only if there exists a horizontal line which intersects the graph of given function at least at two point.

**Explanation by definition**: Let fx=f(y)

⇒ 9-x2=9-y2⇒x2=y2⇒x=±y. But x,y0,3. Therefore, x=-y is not possible (except 0). Thus, x=y. Hence, f is one to one.

**Final statement:** Any horizontal line will intersect the given graph exactly once. Therefore, by using the Horizontal Line test, we can conclude that the given function is one to one.

#### To determine

**b)To find:**

The inverse of f.

#### Answer

-1/2

#### Explanation

First find The inverse of f.

**Step 1**: Let y=f(x)⇒y=9-x2 ⇒x=9-y.

**Step 2**: Interchange the positions of x and y, we get y=9-x.

Thus,f-1x=9-x.

**Calculation of**f-1'8: f'x=-2x. Therefore, f'f-18=f'1=-2, which is non-zero.

So by theorem 7, f-1'8=1f'f-18=1-2.

#### To determine

**c) To calculate:**

f
−1
(x) and its domain and range for f(x)=9−
x
2

#### Explanation

We have already calculated f-1x=9-x in Part (b). The graph for f-1 is as follows:

Domain: Since square root function is defined for non-negative reals, f is defined for 9-x≥0 i.e.,9≥x.

The range of inverse function will be domain of the function.

#### To determine

**d) To calculate:**

(
f
−1
)'(a) for f(x)=9−
x
2

#### Explanation

f-1'x= -129-x

⇒f-1'8= -129-8= -12, which is the same as we calculated in Part (b).

#### To determine

**e) To sketch:**

The graphs of
f and
f
−1
on the same axes for f(x)=9−
x
2

#### Explanation

The orange-colored graph represents the original function f and the blue-coloured graph represents the inverse function of f. Graphs of f and f-1 are symmetrical about the line y=x.