Math Problem Statement

Show that f(x) = x to 3

dt / sqrt(1 + t^4) is one-to-one. f '(x) =

Correct: Your answer is correct. Since f '(x)

Correct: Your answer is correct. 0 for all x, f is increasing

Correct: Your answer is correct. on (−∞, ∞). Thus f is one-to-one. Find (f −1)'(0). (f −1)'(0) =

Incorrect: Your answer is incorrect.

Solution

Let's break this problem down carefully and provide the solution.


Step 1: Understanding f(x)f(x)

The function is defined as: f(x)=x3dt1+t4f(x) = \int_x^3 \frac{dt}{\sqrt{1 + t^4}}


Step 2: Compute f(x)f'(x) using the Fundamental Theorem of Calculus

By the Fundamental Theorem of Calculus, we have f(x)=11+x4.f'(x) = -\frac{1}{\sqrt{1 + x^4}}. Note the negative sign because the variable of integration xx is the lower limit of the integral.

Since 1+x4>01 + x^4 > 0 for all xx, the denominator is always positive. Thus, f(x)<0f'(x) < 0 for all xx. This implies that the function is strictly decreasing on (,)(-\infty, \infty), and hence it is one-to-one.


Step 3: Use the formula for the derivative of the inverse function

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

We are tasked with finding (f1)(0)(f^{-1})'(0). To do this, we need to find x0x_0 such that f(x0)=0f(x_0) = 0.


Step 4: Solve f(x)=0f(x) = 0

x3dt1+t4=0.\int_x^3 \frac{dt}{\sqrt{1 + t^4}} = 0. This implies that x=3x = 3 because the integral of a positive function can only vanish when the upper and lower limits are the same.

Thus, f(3)=0f(3) = 0. So f1(0)=3f^{-1}(0) = 3.


Step 5: Compute (f1)(0)(f^{-1})'(0)

(f1)(0)=1f(3)=111+34=1+81=82.(f^{-1})'(0) = \frac{1}{f'(3)} = \frac{1}{-\frac{1}{\sqrt{1 + 3^4}}} = -\sqrt{1 + 81} = -\sqrt{82}.


Final Answer:

(f1)(0)=82.(f^{-1})'(0) = -\sqrt{82}.

Would you like further clarification or details on any steps?


Related Questions:

  1. How do you apply the Fundamental Theorem of Calculus in general problems?
  2. Why does the sign of f(x)f'(x) determine whether a function is one-to-one?
  3. How is the formula for the derivative of the inverse function derived?
  4. What are conditions for the existence of an inverse for a function?
  5. How can you visualize the monotonicity of a function graphically?

Tip: Always verify whether a function is monotonic before concluding it has an inverse.

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Math Problem Analysis

Mathematical Concepts

Fundamental Theorem of Calculus
Derivative of an Inverse Function
Monotonicity
Integral Calculus

Formulas

f'(x) = -1 / sqrt(1 + x^4)
(f^{-1})'(y) = 1 / f'(f^{-1}(y))
f(x) = ∫(x to 3) dt / sqrt(1 + t^4)

Theorems

Fundamental Theorem of Calculus
Inverse Function Theorem

Suitable Grade Level

Undergraduate