#### To determine

**To find:** (a) is one to one (b) domain and range (c) f-12 (d) Estimated value of f-10

#### Answer

(a) By the horizontal line test, the given function is one-to-one.

(b) Domain and range of f-1 are [-1, 3] and [-3, 3].

(c) f-12=0

(d) f-10≈-1.7

#### Explanation

a) We can see that any horizontal line will intersect the given graph exactly once. Therefore, by using the Horizontal Line Test, we can conclude that the graph for the given function is one to one.

b) To find out domain, look at the x-axis. The given graph starts from x=-3 and ends at x=3, and takes all the values in between. Therefore, the Domain is closed interval

[-3, 3].

To find out range, look at the y-axis. The given graph takes values from -1 to 3 and all values in between. Therefore, the Range is closed interval [-1, 3], as the graph for the given function is bounded by 3 and -1.

The domain of f is the range of f-1 and the range of f is the domain of f-1.

So the domain and range of f-1 are [-1, 3] and [-3, 3].

c) f-1(2) represents the pre-image of 2 under f. With the help of the graph, we can see that f0=2.

The given function is one to one. Therefore,f-12=0.

d)f-10= x

Multiply f on both sides of the equation.

ff-10=f x

Since f and f-1 are cancelled out each other.

f x=0

It is clear from the graph that the preimage of 0 with respect to f is -1.7 approximately.

f-10≈-1.7

**Conclusion:**

(a) By the horizontal line test, the given function is one-to-one.

(b) Domain and range of f-1 are [-1, 3] and [-3, 3]..

(c) f-12=0

(d) f-10≈-1.7