Math Problem Statement

For the functions below, find the tangent line to the inverse of the function at the given x value. You may assume the functions indeed have inverses on the interval [0, ∞). (a) f (x) = x3 + 2x + 5, x = 5

GPT 3.5

Solution

To find the tangent line to the inverse of the function f(x)=x3+2x+5f(x) = x^3 + 2x + 5 at a given xx value, we can follow these steps:

  1. Find yy such that f(y)=5f(y) = 5: This will help us determine the point on the inverse function.

  2. Calculate f(x)f'(x): This will help us find the slope of the tangent line to the inverse function.

  3. Use the slope of the tangent line: The slope of the tangent line to the inverse function at a point is the reciprocal of the derivative of the original function at that point.

Step 1: Solve for yy such that f(y)=5f(y) = 5

We need to solve: y3+2y+5=5y^3 + 2y + 5 = 5 Simplifying this gives: y3+2y=0y^3 + 2y = 0 Factoring out yy: y(y2+2)=0y(y^2 + 2) = 0 This gives y=0y = 0 as a solution. Since y2+2y^2 + 2 has no real roots, we have: f(0)=5f(0) = 5 Thus, f1(5)=0f^{-1}(5) = 0.

Step 2: Calculate f(x)f'(x)

We compute the derivative: f(x)=3x2+2f'(x) = 3x^2 + 2

Step 3: Evaluate f(y)f'(y) at y=0y = 0

Now, we find: f(0)=3(0)2+2=2f'(0) = 3(0)^2 + 2 = 2

Step 4: Find the slope of the tangent line to the inverse

The slope of the tangent line to the inverse function at x=5x = 5 is: slope=1f(y)=12\text{slope} = \frac{1}{f'(y)} = \frac{1}{2}

Step 5: Equation of the tangent line

The equation of a line in point-slope form is: yy1=m(xx1)y - y_1 = m(x - x_1) Where (x1,y1)=(5,0)(x_1, y_1) = (5, 0) and m=12m = \frac{1}{2}. Substituting in: y0=12(x5)y - 0 = \frac{1}{2}(x - 5) This simplifies to: y=12x52y = \frac{1}{2}x - \frac{5}{2}

Conclusion

The equation of the tangent line to the inverse of the function at x=5x = 5 is: y=12x52y = \frac{1}{2}x - \frac{5}{2}

If you have any questions or need further details, feel free to ask! Here are some related questions you might consider:

  1. What is the value of f1(x)f^{-1}(x) for other values of xx?
  2. How can we confirm that the function f(x)f(x) is one-to-one?
  3. What would happen if we changed the function to f(x)=x32x+5f(x) = x^3 - 2x + 5?
  4. How do we graph the function and its inverse?
  5. Can we find the tangent line for another point on the inverse function?

Tip: When working with inverse functions, always ensure that the original function is one-to-one over the interval of interest.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Inverse Functions
Differentiation

Formulas

f'(x) = 3x^2 + 2
y - y_1 = m(x - x_1)

Theorems

Inverse Function Theorem
Point-Slope Form of a Line

Suitable Grade Level

Grades 11-12