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Question

Question: How do you find the inverse of \(y={{x}^{2}}+12x\)?...

How do you find the inverse of y=x2+12xy={{x}^{2}}+12x?

Explanation

Solution

We have the variable xx in more than one term. We will use the completing square method to obtain an equation that has the variable xx in only one term. Then we will rearrange the equation so that we get the value of the variable xx in terms of the variable yy. Then we will get the inverse of the given function by swapping both the variables and replacing the variable yy with f1(x){{f}^{-1}}\left( x \right).

Complete step-by-step solution:
The given equation is y=x2+12xy={{x}^{2}}+12x. Let us modify the given equation so that the variable xx is in only one term. For this, we will use the completing square method to add a constant term in the following manner,
y=x2+12x+(122)2(122)2 y=x2+12x+3636 \begin{aligned} & y={{x}^{2}}+12x+{{\left( \dfrac{12}{2} \right)}^{2}}-{{\left( \dfrac{12}{2} \right)}^{2}} \\\ & \therefore y={{x}^{2}}+12x+36-36 \\\ \end{aligned}
Using the identity (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}, we can write the above equation as the following,
y=(x+6)236y={{\left( x+6 \right)}^{2}}-36
Let us rearrange the above equation so that we obtain the value of the variable xx in terms of the variable yy. We have the following,
(x+6)2=y+36{{\left( x+6 \right)}^{2}}=y+36
Taking square root on both sides, we get
x+6=±y+36 x=6±y+36 \begin{aligned} & x+6=\pm \sqrt{y+36} \\\ & \therefore x=-6\pm \sqrt{y+36} \\\ \end{aligned}
Now, let us swap both the variables in the above equation. We get the following equation,
y=6±x+36y=-6\pm \sqrt{x+36}
Replacing the variable yy with f1(x){{f}^{-1}}\left( x \right), we get the inverse of the given function as follows,
f1(x)=6±x+36{{f}^{-1}}\left( x \right)=-6\pm \sqrt{x+36}

Note: The inverse function is a function that reverses the given function. So, we can plug in the output in the inverse function and obtain the input as the answer. We should be careful with the signs of the terms while shifting them from one side of the equation to the other. Doing the calculation explicitly helps us avoid making any minor errors.