#### To determine

**(a)**

What is a one-to-one function? How can you tell it a function is one-to-one by looking at its graph.

#### Answer

f(x1)≠f(x2) where x1≠x2.

#### Explanation

**Given:** function f is a one-to-one function.

One-to-one function: It is a function in which no two elements of the domain have the same image. In other words, *f* is a one-to-one function if f(x1)≠f(x2) whenever x1≠x2.

For example: f(x)=x3, put the values of x1=1 and x1=2 then f(x1)=1 and f(x2)=8

So, f(x1)≠f(x2).

Graph is a one-to-one function: If *f* is a one-to-one function then no two points (x1,y1),(x2,y2) have the same *y-*value. Therefore, no horizontal line cuts the graph of the equation y=f(x) more than once.

**Conclusion:** Hence, if f(x1)≠f(x2) whenever x1≠x2 the function f is one-to-one function.

#### To determine

**(b)**

If *f* is a one-to-one function, how is its inverse function f−1 defined? How do you obatain the graph of f−1 from the graph of *f* .

#### Answer

Calculated the inverse of the one-to-one function.

#### Explanation

**Given:** function f is a one-to-one function.

To find the inverse of one-to-one function:

1. Replace f(x) with *y*.

2. Inter change *x* and *y*.

3. Solve this equation for *y*. the resulting equation is f−1(x).

For example- f(a)=x3,y=x3 interchange x and y then x=y3 then y=f−1(x)=1x3.

If f is a one-to-one continous function defines on an interval, then inverse f−1 is also one-to-one and continuous function. Thus f−1(x) has an inverse which has to be f(x), by the equivalence of equation given in the defination of the inverse function.

For example-

(f−1)'(a)=1f'(f−1(a))

y=f−1(x) if x=f(y), using implicit differentiable function with inverse function f−1 and f'(f−1(a))≠0

1=f'(y)dydx

Or

1f'(y)=(f−1)'(x)

**Conclusion:** Hence, calculated the inverse of a one-to-one function.

#### To determine

**(c)**

If f is a one-to-one function and f'(f−1(a))≠0, write a fomulae for (f−1)'(a)

#### Answer

(f−1)(a)=1f'(f−1(a)).

#### Explanation

**Given:** f'(f−1(a))≠0.

Let =f−1(x) if and only if x=f(y).

Using implicit differentiation, we differentiate

x=f(y) with respect to x to get

1=f'(y)dydxdydx=1f'(y)1f'(y)=(f−1)'(x)

Replace *x* by *a* for this equation.

(f−1)'(a)=1f'(f−1(x))

**Conclusion:** Hence, if f'(f−1(x))≠0 the formulae of (f−1)'(a)=1f'(f−1(x)).