( 2x+2y=80) , ( y=x+8) 相关知识点: 试题来源: 解析 Find the ( AX=B) from the system of equations. ( [(array)(cc)2& 2 -1& 1(array)]⋅ [(array)cx y(array)]=[(array)c80 8(array)]) Find the inverse of the coefficientmatrix of ( [(array)(cc)2& 2 -1& 1(array)...
This equation does not describe xx as a function of yy because there are two solutions to this equation for every y>0y>0. The problem with trying to find an inverse function for f(x)=x2f(x)=x2 is that two inputs are sent to the same output for each output y>0y>0. The ...
What is the inverse of the function \sqrt {8+2x}? Is it invertible? What is the inverse of the function y = 3x. What is the inverse function for (2x - 3)/4? What is the inverse of the function y = 4x + 5? What is the inverse function of y = \frac{1}{3x+2} ?
Q.3. Find the inverse of the function \(f(x) = 2x – 3\).Ans:Step 1: Substitute \(f(x)\) with \(x\):\(y = 2x – 3\)Step 2: Swap the variables \(x\) and \(y\)\(x = 2y – 3\)Step 3: Solve the equation for \(y\):\(2y = x + 3\)\(y = \frac{{x +...
In order to find the inverse function, we have to first interchange the variables x and y. After interchanging the variables, we need to solve for y. Lastly, we need to simplify.Answer and Explanation: We are given the function y=2x. We want to find the inverse of the fu...
So, the inverse function is $( f^{-1}(x) = \sqrt[3]{x - 2} )$. Example: Inverse of a Rational Function Find the inverse of $( f(x) = \frac{2x + 1}{x - 3} )$. Replace ( f(x) ) with $( y ): ( y = \frac{2x + 1}{x - 3} )$. ...
2x+2y=802x+2y=80 , y=x+8y=x+8 Find the AX=BAX=B from the system of equations. [22−11]⋅[xy]=[808][22-11]⋅[xy]=[808]Find the inverse of the coefficient matrix. Tap for more steps... [14−121412][14-121412]Left multiply both sides of the matrix equation by the ...
An inverse function reverses the method of a function. If, in algebra, f(x) = 2x, then inverse function of f(x) will be a function say f(y) = y/2. Graph, properties, types at BYJU’S
Answer to: Find the inverse: y = \dfrac{x + 8}{x + 2} By signing up, you'll get thousands of step-by-step solutions to your homework questions...
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