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Question: Find the inverse of the quadratic function \(f\left( x \right)=-{{x}^{2}}+2,x\ge 0\) (a) \(\sqrt{\...

Find the inverse of the quadratic function f(x)=−x2+2,x≥0f\left( x \right)=-{{x}^{2}}+2,x\ge 0
(a) (2+x)\sqrt{\left( 2+x \right)}
(b) (2−x)\left( 2-x \right)
(c) (2+x)2\sqrt{{{\left( 2+x \right)}^{2}}}
(d) (2−x)\sqrt{\left( 2-x \right)}

Explanation

Solution

To find the inverse of f(x)=−x2+2f\left( x \right)=-{{x}^{2}}+2 , we have to replace f(x)f\left( x \right) with y. Then, we have to solve for x. We will ignore the negative value since x≥0x\ge 0 . Now, we have to replace x with y. Finally, we have to replace y with f−1(x){{f}^{-1}}\left( x \right) .

Complete step by step answer:
We have to find the inverse of the quadratic function f(x)=−x2+2f\left( x \right)=-{{x}^{2}}+2 . Firstly, we have to replace f(x)f\left( x \right) with y.
⇒y=−x2+2\Rightarrow y=-{{x}^{2}}+2
Let us solve for x. We have to take 2 to the LHS.
⇒y−2=−x2\Rightarrow y-2=-{{x}^{2}}
Now, we have to take the negative sign to the LHS.
⇒−(y−2)=x2 ⇒−y+2=x2 ⇒x2=2−y \begin{aligned} & \Rightarrow -\left( y-2 \right)={{x}^{2}} \\\ & \Rightarrow -y+2={{x}^{2}} \\\ & \Rightarrow {{x}^{2}}=2-y \\\ \end{aligned}
Let us take square roots on both sides.
⇒x=±2−y ⇒x=2−y,−2−y \begin{aligned} & \Rightarrow x=\pm \sqrt{2-y} \\\ & \Rightarrow x=\sqrt{2-y},-\sqrt{2-y} \\\ \end{aligned}
We are given that x≥0x\ge 0 . Therefore, we will ignore the negative value.
⇒x=2−y\Rightarrow x=\sqrt{2-y}
Now, we have to replace x with y.
⇒y=2−x\Rightarrow y=\sqrt{2-x}
We have to replace y with f−1(x){{f}^{-1}}\left( x \right) .
⇒f−1(x)=2−x\Rightarrow {{f}^{-1}}\left( x \right)=\sqrt{2-x}
Therefore, the inverse of f(x)=−x2+2f\left( x \right)=-{{x}^{2}}+2 is 2−x\sqrt{2-x} .

So, the correct answer is “Option d”.

Note: Students must be thorough in solving algebraic equations and the rules involved in it. They should never miss to solve for x in step 2. If so, they have to solve for y in the second last step. After step 1, we will obtain
⇒y=−x2+2\Rightarrow y=-{{x}^{2}}+2
Then, we have to replace x with y.
⇒x=−y2+2,y≥0\Rightarrow x=-{{y}^{2}}+2,y\ge 0
Now, we have to solve for y.
⇒x−2=−y2 ⇒−x+2=y2 ⇒y2=2−x \begin{aligned} & \Rightarrow x-2=-{{y}^{2}} \\\ & \Rightarrow -x+2={{y}^{2}} \\\ & \Rightarrow {{y}^{2}}=2-x \\\ \end{aligned}
Let us take square roots on both sides.
⇒y=±2−x\Rightarrow y=\pm \sqrt{2-x}
We have to ignore the negative value since y≥0y\ge 0 .
⇒y=2−x\Rightarrow y=\sqrt{2-x}
We have to replace y with f−1(x){{f}^{-1}}\left( x \right) .
⇒f−1(x)=2−x\Rightarrow {{f}^{-1}}\left( x \right)=\sqrt{2-x}