Math Problem Statement

solve 29 and 31

solve 29 and 31

Solution

To solve these problems, we need to use the fact that the derivative of the inverse function at a given point aa is:

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

We'll proceed with each question individually.

Problem 29

Given: f(x)=x3+2f(x) = x^3 + 2 and a=1a = -1.

  1. Verify that ff has an inverse:
    Since f(x)=x3+2f(x) = x^3 + 2 is a strictly increasing function (its derivative f(x)=3x2f'(x) = 3x^2 is positive for all x0x \neq 0), ff is one-to-one and hence has an inverse.

  2. Find f1(a)f^{-1}(a): We need to solve for xx such that f(x)=1f(x) = -1: x3+2=1x^3 + 2 = -1 x3=3x^3 = -3 x=33x = -\sqrt[3]{3} So, f1(1)=33f^{-1}(-1) = -\sqrt[3]{3}.

  3. Find f(x)f'(x): f(x)=3x2f'(x) = 3x^2

  4. Evaluate f(f1(a))f'(f^{-1}(a)): Substitute x=33x = -\sqrt[3]{3} into f(x)f'(x): [ f'(f^{-1}(-1)) = f'(-\sqrt[3]{3}) = 3(-\sqrt[3]{3})^2 = 3 \cdot \sqrt[3]{9} ]

  5. Calculate (f1)(1)(f^{-1})'(-1): (f1)(1)=1393(f^{-1})'(-1) = \frac{1}{3 \cdot \sqrt[3]{9}}

So, the answer for problem 29 is: (f1)(1)=1393(f^{-1})'(-1) = \frac{1}{3 \cdot \sqrt[3]{9}}


Problem 31

Given: f(x)=tanx,π4xπ4f(x) = \tan x, \quad -\frac{\pi}{4} \leq x \leq \frac{\pi}{4} and a=33a = \frac{\sqrt{3}}{3}.

  1. Verify that ff has an inverse:
    Within the interval π4xπ4-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}, f(x)=tanxf(x) = \tan x is strictly increasing and continuous, thus it is one-to-one and has an inverse.

  2. Find f1(a)f^{-1}(a): We need to solve for xx such that f(x)=33f(x) = \frac{\sqrt{3}}{3}: tanx=33\tan x = \frac{\sqrt{3}}{3} The solution within the given interval is: x=π6x = \frac{\pi}{6} Thus, f1(33)=π6f^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}.

  3. Find f(x)f'(x): f(x)=sec2xf'(x) = \sec^2 x

  4. Evaluate f(f1(a))f'(f^{-1}(a)): Substitute x=π6x = \frac{\pi}{6} into f(x)f'(x): f(f1(33))=f(π6)=sec2(π6)f'\left(f^{-1}\left(\frac{\sqrt{3}}{3}\right)\right) = f'\left(\frac{\pi}{6}\right) = \sec^2\left(\frac{\pi}{6}\right) Since sec(π6)=23\sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}, sec2(π6)=(23)2=43\sec^2\left(\frac{\pi}{6}\right) = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}

  5. Calculate (f1)(33)\left(f^{-1}\right)'\left(\frac{\sqrt{3}}{3}\right): (f1)(33)=143=34\left(f^{-1}\right)'\left(\frac{\sqrt{3}}{3}\right) = \frac{1}{\frac{4}{3}} = \frac{3}{4}

So, the answer for problem 31 is: (f1)(33)=34\left(f^{-1}\right)'\left(\frac{\sqrt{3}}{3}\right) = \frac{3}{4}


Would you like a further breakdown of these steps, or do you have any other questions?

Related Questions

  1. How can we generalize the formula for the derivative of an inverse function?
  2. Why is the domain restriction necessary for f(x)=tanxf(x) = \tan x to have an inverse?
  3. Can you explain why a function being strictly increasing implies it has an inverse?
  4. How would the derivative of the inverse change if the function were decreasing instead of increasing?
  5. What would be the steps to find the inverse derivative of a composite function?

Tip

When verifying if a function has an inverse, checking if it is strictly monotonic (either increasing or decreasing) over the entire domain can simplify the process.

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Math Problem Analysis

Mathematical Concepts

Inverse Functions
Derivatives
Trigonometry

Formulas

Derivative of an inverse function: (f^{-1})'(a) = 1 / f'(f^{-1}(a))
Derivative of cubic functions
Derivative of the tangent function: f'(x) = sec^2(x)

Theorems

Theorem on the derivative of inverse functions (Theorem 5.9)

Suitable Grade Level

Grades 11-12