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Show that the function \(f\) has an inverse function, and find \(\left[D_{x} f^{-1}(x)\right]_{x=a}\) for the given number \(a\). $$ f(x)=2 x^{3}-8 x+5, \quad-1 \leq x \leq 1 ; \quad a=5 $$

Short Answer

Expert verified
Yes, the inverse exists, and \([D_x f^{-1}(x)]_{x=5} = -\frac{1}{8}.\)

Step by step solution

01

Verify the Injectivity of the Function

To have an inverse, the function needs to be bijective (injective and surjective) on the given domain. To check injectivity on the interval \([-1, 1]\), calculate the derivative and show it's always positive or always negative. \[ f'(x) = \frac{d}{dx}(2x^3 - 8x + 5) = 6x^2 - 8. \] Evaluate \(f'(x)\) at a few points within the interval. If there is no change in sign, the function is injective.
02

Evaluate the Derivative for Monotonicity

Check for monotonicity by assessing the sign of \(f'(x)\):1. \(f'(-1) = 6(-1)^2 - 8 = -2.\)2. \(f'(0) = -8.\)3. \(f'(1) = 6(1)^2 - 8 = -2.\)The derivative is negative across the interval \([-1, 1]\), indicating that \(f(x)\) is strictly decreasing and, therefore, injective.
03

Verify Surjectivity

Since \(f(x)\) is injective and continuous on \([-1,1]\), check that it reaches both the maximum and minimum within this interval. Compute \(f(-1)\) and \(f(1)\):1. \(f(-1) = 2(-1)^3 - 8(-1) + 5 = -2 + 8 + 5 = 11.\)2. \(f(1) = 2(1)^3 - 8(1) + 5 = 2 - 8 + 5 = -1.\)The continuous function ranges from \(11\) to \(-1\), covering the output range of \([-1,1]\), ensuring surjectivity.
04

Find the Inverse Function's Derivative Formula

The derivative of the inverse function at a specific point is expressed as \(\left[D_x f^{-1}(x)\right]_{x=a} = \frac{1}{f'(f^{-1}(a))}.\)
05

Solve for \(f^{-1}(a)\)

Find \(x\) such that \(f(x) = a = 5\). We have \(f(x) = 2x^3 - 8x + 5.\) Plugging in \(f(x) = 5\):\(2x^3 - 8x + 5 = 5 \2x^3 - 8x = 0 \x(2x^2 - 8) = 0.\)Solve for \(x\):1. \(x = 0\)2. \(2x^2 - 8 = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2\) or \(x = -2\), but \(x = 0\) is within our domain (\([-1, 1]\)).
06

Compute the Derivative at the Inverse Function Value

Using \(f^{-1}(5) = 0\), now substitute into \(\left[\frac{1}{f'(f^{-1}(5))}\right]\):1. Calculate \(f'(0)\), which is already determined to be \(-8\). 2. \[ \left[D_x f^{-1}(x)\right]_{x=5} = \frac{1}{-8} = -\frac{1}{8}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Injective Function
An injective function, also known as a one-to-one function, is a function where each element of the domain maps to a unique element in the codomain. This means no two different inputs can have the same output. To determine if a function is injective, we often use the derivative test on its domain.

If the derivative is always positive or always negative over that range, the function is monotonic, and hence injective. In the given problem, we calculate the derivative of the function:
  • Given function: \( f(x) = 2x^3 - 8x + 5 \)
  • Derivative: \( f'(x) = 6x^2 - 8 \)
By evaluating this derivative at various points within the interval
  • \([-1, 1]\)
we saw that \( f'(x) \) is negative throughout the interval, confirming the function is strictly decreasing and thereby injective.
Surjective Function
A surjective function, also called an onto function, is one where every element in the codomain is mapped by some element in the domain. To check for surjectivity, especially in a continuous function, we can evaluate the function at the endpoints of its domain and see if it covers the entire output interval.

In the exercise, we computed:
  • \( f(-1) = 11 \)
  • \( f(1) = -1 \)
This shows that the function spans from 11 to -1 across the interval \([-1, 1]\). Since the function is continuous, it must cover all values between its maximum and minimum, verifying that it is surjective. Therefore, the injective and surjective properties together ensure that the function is bijective, allowing it to have an inverse.
Derivative
The derivative of a function is a measure of how the function value changes as its input changes. It indicates the rate of change or the slope of the tangent line to the function at any given point. In this problem, the derivative of the given function:
  • \( f(x) = 2x^3 - 8x + 5 \)
was calculated as:
  • \( f'(x) = 6x^2 - 8 \)
Evaluating the derivative at various points within the domain helps us check monotonicity and determine injectivity. Additionally, for inverse functions, the derivative gives us insight into how rapidly the inverse is changing at any point by using the relationship:
  • \( \left[D_x f^{-1}(x)\right]_{x=a} = \frac{1}{f'(f^{-1}(a))} \)
In this exercise, since we solved \( f(x) = a \) and found \( x=0 \) for \( a=5 \), the derivative of the inverse function at \( x=5 \) was computed as \(-\frac{1}{8}\).
Monotonicity
Monotonicity refers to a function either always increasing or always decreasing on a given interval. This property makes a function easily invertible because it means the function is either strictly increasing or strictly decreasing, ensuring bijectivity on that interval.

For the function \( f(x) = 2x^3 - 8x + 5 \), we used its derivative:
  • \( f'(x) = 6x^2 - 8 \)
to investigate monotonicity. By checking the sign of the derivative within the range \([-1, 1]\), and finding it always negative, we verified that the function is strictly decreasing.

This continual decrease implies that no matter which two values are chosen within this domain, the function will consistently order itself. As a result, this property aids in proving that the function has an inverse.

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