# If g^-1 is the inverse function of g, write an equation for the line tangent to the graph of y = g^-1(x) at x =2. Given: g(1) =2 and g'(1) =5

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### 2 Answers

You need to use the following equation that relates the derivative of a function and its inverse, such that:

`g'(x)*(g^(-1)(x))' = 1 => (g^(-1)(x))' = 1/g'(x)`

The problem provides the information `g'(1) = 5` , hence, replacing 5 for `g'(1)` in equation above yields:

`(g^(-1)(x))' = 1/5`

You need to evaluate the equation of the tangent line to the curve `g^(-1)(x) = y` , at `x = 2` , such that:

`y - g^(-1)(2) = 1/5(x - 2)`

Since the problem provides the information that `g(1) = 2` , hence, `g^(-1)(2) = 1`

`y - 1 = 1/5(x - 2) => y = x/5 - 2/5 + 1 => y = x/5 + 3/5`

**Hence, evaluating the equation of tangent line to the curve `g^(-1)(x) = y` , at `x = 2` , yields `y = x/5 + 3/5.` **

The equation of the line tangent to the graph of `y = g^-1(x)` at x = 2 has to be determined.

`y = g^-1(x)`

=> `x = g(y)`

determine `dy/dx` using implicit differentiation

`1 = g'(y)*(dy/dx)`

=> `dy/dx = 1/(g'(y))`

As `g(y) = x` and `g(1) = 2` , for x = 2, y = 1

At x = 2, g'(1) = 5

The slope of the required tangent is `1/(g'(1)) = 1/5` . The equation of the line is `(y - 1)/(x - 2) = 1/5`

=> 5y - 5 = x - 2

=> x - 5y + 3 = 0

**The equation of the required tangent is x - 5y + 3 = 0**