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)).