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Consider f: R^{+} \rightarrow[-5, \infty) given by f(x)=9 x^{2}+6 x-5 . Show that f is invertible with f^{-1}(x)=(v(x+6)-1) / 3
CBSE | Class 12 | Exercise 2.4 | Function | Maths | RD Sharma
Consider f: R^{+} \rightarrow[-5, \infty) given by f(x)=9 x^{2}+6 x-5 . Show that f is invertible with f^{-1}(x)=(v(x+6)-1) / 3

Given f: R^{+} \rightarrow[-5, \infty) given by f(x)=9 x^{2}+6 x-5 We have to show that \mathrm{f} is invertible.

\section{Injectivity of f:}

Let x and y be two elements of domain \left(R^{+}\right),

Such that f(x)=f(y)

\Rightarrow 9 x^{2}+6 x-5=9 y^{2}+6 y-5

\Rightarrow 9 x^{2}+6 x=9 y^{2}+6 y

\Rightarrow x=y\left(A s, x, y \in R^{+}\right)

So, f is one-one.

Surjectivity of f:

Let y is in the co domain (Q)

Such that f(x)=y

\Rightarrow 9 x^{2}+6 x-5=y

\Rightarrow 9 x^{2}+6 x=y+5

\Rightarrow 9 x^{2}+6 x+1=y+6 (By adding 1 on both sides )

\Rightarrow(3 x+1)^{2}=y+6

\Rightarrow 3 x+1=v(y+6)

\Rightarrow 3 x=v(y+6)-1

\Rightarrow x=(V(y+6)-1) / 3 in R^{+}(domain )

\mathrm{f} is onto.

So, \mathrm{f} is a bijection and hence, it is invertible.

Now we have to find \mathrm{f}^{-1}

Let f^{-1}(x)=y \ldots .

\Rightarrow x=f(y)

\Rightarrow x=9 y^{2}+6 y-5

\Rightarrow x+5=9 y^{2}+6 y

\Rightarrow x+6=9 y^{2}+6 y+1 \quad (adding 1 on both sides)

\Rightarrow x+6=(3 y+1)^{2}

\Rightarrow 3 y+1=v(x+6)

\Rightarrow 3 y=V(x+6)-1

\Rightarrow y=(v(x+6)-1) / 3

Now substituting this value in (1) we get,

So, f^{-1}(x)=(V(x+6)-1) / 3

i

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