If $y=4 x^{2}-16$, what is its inverse?
$y^{-1}=-\sqrt{\frac{x}{4}}+16$ such that $x \geq 0$
$y^{-1}=-\sqrt{\frac{x}{4}}+16$ such that $x \leq 0$
$y^{-1}=-\sqrt{\frac{x+16}{4}}$ such that $x \leq-16$
$x^{-1}=-\sqrt{\frac{x+16}{4}}$ such that $x \geq-16$
Answer
\(\boxed{y=\pm\sqrt{\frac{x+16}{4}}}\) is the inverse of the function \(y=4x^{2}-16\). This means that for any \(x\) in the domain of the original function, \(y^{-1}(x)\) will give the corresponding \(y\) value in the original function. Note that the \(\pm\) indicates that there are two possible values for \(y^{-1}(x)\), one positive and one negative. This is because the original function is a quadratic, which can have two different \(x\) values for a given \(y\) value.
Step 1 ：Given the function \(y=4x^{2}-16\), we are asked to find its inverse.
Step 2 ：First, we swap the roles of \(x\) and \(y\) in the equation, which gives us \(x=4y^{2}-16\).
Step 3 ：Now, we solve this equation for \(y\).
Step 4 ：Adding 16 to both sides of the equation, we get \(x+16=4y^{2}\).
Step 5 ：Dividing both sides by 4, we get \(\frac{x+16}{4}=y^{2}\).
Step 6 ：Taking the square root of both sides, we get \(y=\pm\sqrt{\frac{x+16}{4}}\).
Step 7 ：\(\boxed{y=\pm\sqrt{\frac{x+16}{4}}}\) is the inverse of the function \(y=4x^{2}-16\). This means that for any \(x\) in the domain of the original function, \(y^{-1}(x)\) will give the corresponding \(y\) value in the original function. Note that the \(\pm\) indicates that there are two possible values for \(y^{-1}(x)\), one positive and one negative. This is because the original function is a quadratic, which can have two different \(x\) values for a given \(y\) value.
