Math Problem Statement

Find the slope of the line tangent to the inverse function at the point P(1, 1): and find the equation of the tangent line to the inverse function.

Solution

The function given is:

y=f(x)=65+x2y = f(x) = \frac{6}{5 + x^2}

Part a: Finding the slope of the line tangent to the inverse function at the point P(1,1)P(1, 1).

The slope of the tangent line to the inverse function at a point can be found using the formula for the derivative of the inverse function:

(f1)(a)=1f(f1(a))\left( f^{-1} \right)'(a) = \frac{1}{f'(f^{-1}(a))}

For P(1,1)P(1, 1), since f(1)=1f(1) = 1 (meaning f1(1)=1f^{-1}(1) = 1), we need to find f(1)f'(1).

First, differentiate f(x)f(x) using the quotient rule:

f(x)=(5+x2)(0)(6)(2x)(5+x2)2=12x(5+x2)2f'(x) = \frac{(5 + x^2)(0) - (6)(2x)}{(5 + x^2)^2} = \frac{-12x}{(5 + x^2)^2}

Now, evaluate f(1)f'(1):

f(1)=12(1)(5+12)2=1236=13f'(1) = \frac{-12(1)}{(5 + 1^2)^2} = \frac{-12}{36} = -\frac{1}{3}

The slope of the inverse function's tangent line at P(1,1)P(1, 1) is:

(f1)(1)=1f(1)=113=3\left( f^{-1} \right)'(1) = \frac{1}{f'(1)} = \frac{1}{-\frac{1}{3}} = -3

So, the slope m=3m = -3.

Part b: Finding the equation of the line tangent to the inverse function at P(1,1)P(1, 1).

The equation of a line is given by the point-slope form:

yy1=m(xx1)y - y_1 = m(x - x_1)

Here, the slope m=3m = -3 and the point is P(1,1)P(1, 1). Substituting these values:

y1=3(x1)y - 1 = -3(x - 1)

Simplifying:

y1=3x+3y - 1 = -3x + 3

y=3x+4y = -3x + 4

Thus, the equation of the tangent line is:

y=3x+4y = -3x + 4


Would you like more details or have any further questions?

Related questions:

  1. What is the geometric meaning of the slope of the inverse function's tangent line?
  2. How do you apply the quotient rule to find the derivative of rational functions?
  3. Why does the derivative of the inverse function depend on the derivative of the original function?
  4. What is the significance of the point P(1,1)P(1,1) in the context of the function and its inverse?
  5. How can you check if a function is invertible at a given point?

Tip: The relationship between a function and its inverse can often be understood by switching xx and yy and solving for one variable in terms of the other.

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Math Problem Analysis

Mathematical Concepts

Inverse Functions
Derivatives
Tangent Lines

Formulas

f'(x) = (5 + x^2)(0) - (6)(2x) / (5 + x^2)^2
f^-1(a)' = 1 / f'(f^-1(a))
Point-slope form: y - y_1 = m(x - x_1)

Theorems

Derivative of Inverse Function Theorem
Quotient Rule for Differentiation

Suitable Grade Level

Undergraduate Level (Calculus)