Question

How do you find the inverse of the function $ g(x) = 3x - 5 $ ?

Answer 1

Hint: Since it is a linear equation with no denominator so there is no limitations of domain or range, it can be any real number till $ \infty $ . We have to find $ {g^{ - 1}}(x) $ for the given equation. We know that $ f(x) = y $ so put this value in the given function then solve further to get $ x $ in terms of $ y $ . So equate accordingly.

Complete step-by-step answer:
We are given with the function $ g(x) = 3x - 5 $ . Since it has no denominator so the range and domain can be any real number till infinity.
We know that $ f(x) = y $ so let $ g(x) $ be $ y $ . So, we can write the function as $ y = 3x - 5 $ .
Adding both the sides by $ 5 $ we get:
 $
  y = 3x - 5 \\
  y + 5 = 3x - 5 + 5 \\
  y + 5 = 3x \;
  $
Dividing both the sides by $ 3 $ we get:
 $
  y + 5 = 3x \\
  \dfrac{{y + 5}}{3} = \dfrac{{3x}}{3} \\
  \dfrac{{y + 5}}{3} = x \;
  $ ……………….(i)
Since, we know that $ f(x) = y $ that implies $ {f^{ - 1}}(y) = x $ :
Put this value in the equation (i) and we get:
 $
  \dfrac{{y + 5}}{3} = x \\
  \dfrac{{y + 5}}{3} = {f^{ - 1}}(y) \;
  $
Since, we got it for $ y $ and we need $ x $ . So replace all values of $ y $ from the equation with $ x $ and we get:
 $ \dfrac{{x + 5}}{3} = {f^{ - 1}}(x) $ which is the inverse of the function $ f(x) = 3x - 5 $ .
Similarly, for this function $ g(x) = 3x - 5 $ , the inverse is $ {g^{ - 1}}(x) = \dfrac{{x + 5}}{3} $ .
So, the correct answer is “ $ \dfrac{{x + 5}}{3} $ .”.

Note: Since we know that $ fo{f^{ - 1}}(x) = x $ , so it should be the same for $ go{g^{ - 1}}(x) = x $ .
Let’s check it out for $ go{g^{ - 1}}(x) = x $ .
 $ go{g^{ - 1}}(x) = x $ can be written as $ g({g^{ - 1}}(x)) = x $ .
Solve for left hand side to prove it same as right hand side:
L.H.S $ = g({g^{ - 1}}(x)) $
Put the value of $ {g^{ - 1}}(x) $ above and we get:
 $ = g(\dfrac{{x + 5}}{3}) $
Now put the value of $ \dfrac{{x + 5}}{3} $ in $ g(x) $ and we get:
 $
  3\left( {\dfrac{{x + 5}}{3}} \right) - 5 \\
  x + 5 - 5 \\
  x \;
  $
Which is equal to R.H.S.
Therefore, the inverse $ {g^{ - 1}}(x) = \dfrac{{x + 5}}{3} $ for the given function $ g(x) = 3x - 5 $ is correct.