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Question

Show that the function f:RR defined by f(x)=3x12,xR is one-one and onto functions. Also, find the inverse of the function f.

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Solution

Given,
f(x)=3x12,xR

For one-one :
Let x1,x2R such that f(x1)=f(x2)
3x112=3x212

3x11=3x21
3x1=3x2
x1=x2
Therefore, f is one-one.

For onto :
Let yR, then f(x)=y
3x12=y

3x1=2y
x=2y+13R

Thus, for each yR, there exists xR such that
f(2y+13)=y

Hence, f is onto.
Therefore, f is bijective. Hence, f1 exists.
Now,
x=2y+13
f1(y)=2y+13
Therefore,
f1(x)=2x+13

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