Verify that the function g(x)=x-2/5 is the inverse function of f(x)=5x+2.........5 stars best asnwers?

Question

STEP 1: the function notation f(x) can be rewritten as________(x or y choose the letter that correctly fills in the blank). rewrite the function f(x)=5x+2 with the letter you chose above to replace the function notation.

STEP 2: rewrite the equation you wrote in step 1 with the variable reversed, and solve for y. show your work.

STEP 3: verify that the function g(x)=x-2/5 is the inverse function of f(x)=5x+2. write your answer in inverse notation.

Answer 1

STEP 1: the function notation f(x) can be rewritten as y. y = 5x + 2



STEP 2: x = 5y + 2



.............5y = x - 2



..............y = (x - 2)/5



STEP 3:To verify, one must do f((f^-1)(x)) where (f^-1)(x) is the inverse.



............f((f^-1)(x)) = 5((x - 2)/5) + 2



........... f((f^-1)(x)) = x.....if you get x, then it is the inverse

Answer 2

Inverse Function Notation

Answer 3

right here... once you desire to locate an inverse function, you turn the variables 'x' and 'y' and resolve for y. fairly, f(x) and g(x) are different notations for y, so we in truth have 2 equations: #a million: y = 2x + a million #2: y = (x - a million) / 2 we could desire to instruct that #2 is the inverse of #a million, so we swap the variables in #a million and resolve for y, showing that it is going to likely be comparable to equation #2: x = 2y + a million x - a million = 2y (x - a million) / 2 = y So, we see that the inverse of the #a million is #2. even if, you have merely been waiting to locate the inverse, however the classic verification is to plug the inverse function into the unique and have the equation simplify to easily x, and vica versa. In different words, f(g(x)) = x AND g(f(x)) = x So, take the inverse function and plug it in for x interior the unique function: f(g(x)) = 2((x-a million) /2 ) + a million f(g(x)) = (x-a million) + a million ==> 2's cancel out f(g(x)) = x - a million + a million ==> subtracting and including a million cancels f(g(x)) = x Now, we could desire to plug the unique into the inverse: g(f(x)) = ((2x-a million)+a million)/2 g(f(x)) = ((2x - a million) + a million) / 2 ==> a million's cancel out g(f(x)) = (2x) / 2 ==> 2's cancel out g(f(x)) = x So, now you have efficiently verified it.