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Let R_{+}be the set of all positive real numbers. show that the function f: R_{4} \rightarrow[-5, \infty]: f(x)=\left(9 x^{2}+6 x-5\right) is invertible. Find f^{-1}.
CBSE | Class 12 | Exercise 2D | Functions | Maths | RS Aggarwal
Let R_{+}be the set of all positive real numbers. show that the function f: R_{4} \rightarrow[-5, \infty]: f(x)=\left(9 x^{2}+6 x-5\right) is invertible. Find f^{-1}.

Solution:

f(x)=9 x_{2}+6 x-5 \text { (as given) }
\mathrm{f}(\mathrm{x}) is invertible if \mathrm{f}(\mathrm{x}) is a bijection (i.e one-one onto function)
One-One function
Suppose p,q be two arbitrary elements in \mathrm{R}
Therefore, f(p)=f(q)
\begin{array}{l} \Rightarrow 9 p_{2}+6 p-5=9 q_{2}+6 q-5 \\ \Rightarrow 9 p_{2}+6 p=9 q_{2}+6 q \\ \Rightarrow 9 p_{2}-9 q_{2}+6 p-6 q=0 \\ \Rightarrow 9(p-q)(p+q)+6(p-q)=0 \\ \Rightarrow(p-q)[9(p+q)+6]=0 \\ \Rightarrow p-q=0 \quad \text { or } \quad 9(p+q)+6=0 \\ \Rightarrow p=q \end{array}
When f(p)=f(q), p=q
So, \mathrm{f}(\mathrm{x}) is one-one function.
Onto function
Suppose \mathrm{v} be an arbitrary element of \mathrm{R} (Co-domain)
Therefore, \mathrm{f}(\mathrm{x})=\mathrm{v}
9 \mathrm{x}_{2}+6 \mathrm{x}-5=\mathrm{v}
On Solving, we obtain
\mathrm{x}=\frac{-1+\sqrt{v+6}}{3}
As \mathrm{v} \in \mathrm{R} \rightarrow[-5, \infty]
\Rightarrow \frac{-1+\sqrt{v+6}}{3} \in \mathrm{R} \rightarrow[-5, \infty]
Thus, Range of \mathrm{f}(\mathrm{x})= co-domain of \mathrm{f}(\mathrm{x})=[-5, \infty]
Therefore, \mathrm{f}(\mathrm{x}) is onto function.
As a result, \mathrm{f}(\mathrm{x}) is invertible.
Now we need to find \mathrm{f}_{-1},
Suppose f(x)=y
y=9 x 2+6 x-5
Now, replace all x with y and all y with x.
Now, solve for y
\Rightarrow \mathrm{x}=9 \mathrm{y}_{2}+6 \mathrm{y}-5
On Solving, we obtain
y=\frac{-1+\sqrt{x+6}}{3}
Now replace \mathrm{y} with \mathrm{f}-1(\mathrm{x})
As a result, f_{-1}(x)=\frac{-1+\sqrt{x+6}}{3}

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