Consider a function f(x). If a tangent line is drawn to f(x) at a point (x_{1}, y_{1}), then the equation of the tangent line is
y - y_{1} = m(x - x_{1})
Here, m represents the slope of the tangent line.
To find the slope of the tangent line m, you have to find the first derivative of f(x), that is f'(x). Then, find the value of f'(x) at (x_{1}, y_{1}).
m = value of f'(x) at (x_{1}, y_{1})
Consider the function f(x). Let g(x) be the inverse function of f(x).
g(x) = f^{-1}(x)
To find the slope of a tangent line to g(x), we have to find the first derivative of g(x), that is g'(x). Then, find the value of g'(x) at the point of tangency.
Since g(x) is the inverse of f(x), we can find the derivative of g(x) as follows.
g(x) = f^{-1}(x)
Take the function f on both sides using the operation function composition.
f ∘ g(x)] = f ∘ f^{-1}(x)
f[g(x)] = f[f^{-1}(x)]
f[g(x)] = f^{1}^{-1}(x)
f[g(x)] = f^{0}(x)
f[g(x)] = x
Find the derivative on both sides with respect to x. (Use chain rule on the left side. That is, first find the derivative of f, then by chain rule, find the derivative of g(x)).
f'[g(x)] ⋅ g'(x) = 1
Divide both sides by f'[g(x)].
Example :
Consider the following function.
f(x) = x^{3} + x + 5
Let g(x) be the inverse of f(x). Find the equation of a tangent line to g(x) at x = 5.
Solution :
Since g(x) is the inverse of f(x),
g(x) = f^{-1}(x)
Formula to find the derivative of g(x).
Substitute x = 5.
Let g(5) = k.
Since g(x) is the inverse of f(x),
f(k) = 5
k^{3} + k + 5 = 5
Subtract 5 from both sides.
k^{3} + k = 0
k(k^{2} + 1) = 0
k = 0 |
k^{2} + 1 = 0 √k^{2} = √-1 k = √-1 (imaginary) |
Therefore,
k = 0
Since g(5) =k,
g(5) = 0
So, the point on g(x) corresponding to x = 5 is (5, 0).
f(x) = x^{3} + x + 5
f'(x) = 3x^{2} + 1
Substitute x = 0.
f'(0) = 3(0)^{2} + 1
f'(0) = 0 + 1
f'(0) = 1
Therefore,
g'(5) = ¹⁄₁
g'(5) = 1
Slope of the tangent line to g(x) at x = 5 is 1.
Equation of the tangent line to g(x) :
y - y_{1} = m(x - x_{1})
Substitute (x_{1}, y_{1}) = (5, 0) and m = 1.
y - 0 = 1(x - 5)
y = x - 5
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Oct 01, 24 12:04 PM
Oct 01, 24 11:49 AM
Sep 30, 24 11:38 AM