Question

The graph of \(f\) is given.

(a) Why is \(f\)one-to-one?

(b) What are the domain and range of \({f^{ - 1}}\)?

(c) What is the value of \({f^{ - 1}}\left( 2 \right)\)?

(d) Estimate the value of \({f^{ - 1}}\left( 0 \right)\)?

Short answer

a) \(f\) is a one-to-one function.

b) The range of \({f^{ - 1}}\) is [-3,3] and domain of \({f^{ - 1}}\) is [-1,3].

c) The value of \({f^{ - 1}}\left( 2 \right)\) is 0.

d) The value of \({f^{ - 1}}\left( 0 \right)\) is \( - 1.7\).

Step-by-step solution

Horizontal Line Test

If no horizontal line touches the graph of the function more than once, the function is one-to-one.

Determine whether \(f\) is one-to-one

a)

Here, \(f\) is a one-to-one function since it follows the Horizontal Line Test.

Determine the domain and range of \({f^{ - 1}}\)

b)

It is known thatdomainof \({f^{ - 1}} = \)range\({\mathop{\rm of}\nolimits} \,\,f\) and range of \({f^{ - 1}} = \)domain \({\mathop{\rm of}\nolimits} \,\,f\).

The domain of the function \(f\) \( = \left[ { - 3,3} \right] = \)Range of \({f^{ - 1}}\).

The range of \(f = \left[ { - 1,3} \right] = \)Domain of \({f^{ - 1}}\).

Determine the value of \({f^{ - 1}}\left( 2 \right)\) and \({f^{ - 1}}\left( 0 \right)\)

c)

\({f^{ - 1}}\left( 2 \right) = 0\)because \(f\left( 0 \right) \approx 2\) and \(f\) is a one-to-one function.

Thus, the value of \({f^{ - 1}}\left( 2 \right)\) is 0.

d)

\({f^{ - 1}}\left( 0 \right) = - 1.7\)because \(f\left( { - 1.7} \right) \approx 0\) and \(f\) is a one-to-one function.

Thus, the value of \({f^{ - 1}}\left( 0 \right)\) is \( - 1.7\).