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Chapter 3: Problem 2

Let \(f(x)=x^{3}+2 e^{x}\). (a) Show that \(f\) is one-to-one and confirm that \(f(0)=2\). (b) Find \(\left(f^{-1}\right)^{\prime}(2)\).

Short Answer

Expert verified
(f is one-to-one, f(0)=2, and (f^{-1})鈥�(2)=1/2).

Step by step solution

01

Determine if f is One-to-One

A function is one-to-one if its derivative is either always positive or always negative. Find the derivative of \(f(x)\):\[f'(x) = 3x^2 + 2e^x\]Since \(3x^2\) is always non-negative and \(2e^x\) is always positive, \(f'(x) > 0\) for all \(x\). Hence, \(f(x)\) is strictly increasing, indicating that \(f(x)\) is one-to-one.
02

Verify that f(0) = 2

Substitute \(x = 0\) into \(f(x)\):\[f(0) = 0^3 + 2e^0 = 0 + 2 = 2\]Thus, \(f(0) = 2\) is confirmed.
03

Use the Inverse Function Theorem

By the Inverse Function Theorem, \[(f^{-1})'(b) = \frac{1}{f'(a)}\]where \(f(a) = b\). We know \(f(0) = 2\), so \(a = 0\) and \(b = 2\). We need to find \(f'(0)\) first.
04

Calculate f'(0)

Substitute \(x=0\) into \(f'(x)\):\[f'(x) = 3x^2 + 2e^x\]\[f'(0) = 3(0)^2 + 2e^0 = 0 + 2 = 2\]So, \(f'(0) = 2\).
05

Find (f^{-1})鈥�(2)

Now apply the Inverse Function Theorem:\[(f^{-1})'(2) = \frac{1}{f'(0)} = \frac{1}{2}\]Hence, \((f^{-1})'(2) = \frac{1}{2}\).

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Key Concepts

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

One-to-One Function
A one-to-one function is a special type of function where each input has a unique output. This means no two different inputs map to the same output. In mathematical terms, if \(f(a) = f(b)\), then it must be true that \(a = b\). To determine if the function \(f(x) = x^{3}+2 e^{x}\) is one-to-one, we can use the concept of the derivative. If the derivative of a function, \(f'(x)\), is either always positive or always negative across its domain, the function is strictly increasing or strictly decreasing. This ensures that it is one-to-one.For our function:
  • Find the derivative: \(f'(x) = 3x^2 + 2e^x\).
  • Notice that the term \(3x^2\) is always non-negative and \(2e^x\) is always positive.
  • Thus, \(f'(x) > 0\) for all \(x\), meaning the function is strictly increasing.
Since the derivative is positive everywhere, \(f(x)\) is a one-to-one function.
Derivative
The derivative of a function at a point provides the rate at which the function's value changes as its input changes. It is a core concept in calculus that measures the slope or steepness of the function at any given point. For the function \(f(x) = x^{3}+2 e^{x}\), the derivative \(f'(x)\) can be computed as follows:
  • For the term \(x^3\), the derivative is \(3x^2\) (power rule).
  • For the exponential term \(2e^x\), the derivative is \(2e^x\) (since the derivative of \(e^x\) is \(e^x\)).
Putting it all together, we get \(f'(x) = 3x^2 + 2e^x\). This derivative helps us not only to find if the function is one-to-one but also to apply the inverse function theorem. Especially when finding the derivative at specific points, like \(f'(0)\) in this problem, it indicates the rate of change of the function at that specific point.
Exponential Function
The exponential function is one of the most important mathematical functions, represented as \(e^x\), where \(e\) is Euler's number (approximately 2.71828). Exponential functions have a unique property where the rate of growth or decay of the function is proportional to its current value. In our given function \(f(x) = x^3 + 2e^x\), the term \(2e^x\) plays a significant role:
  • The exponential component \(e^x\) grows rapidly as \(x\) increases, contributing to making \(f(x)\) a one-to-one function since \(2e^x\) is always positive.
  • The presence of \(2e^x\) ensures that the overall function is increasing, even when the \(x^3\) component is small or zero.
The derivative of the exponential function, \(e^x\), is itself, which makes it powerful for analytic processes. This constancy in growth rate helps simplify calculations like finding derivatives, and remains unchanged in behavior across its domain.

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Most popular questions from this chapter

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There is a myth that circulates among beginning calculus students which states that all indeterminate forms of types \(0^{0}, \infty^{0}\), and \(1^{*}\) have value 1 because "anything to the zero power is 1 " and \({ }^{*} 1\) to any power is \(1 . "\) The fallacy is that \(0^{0}, \infty^{0}\), and \(1^{\infty}\) are not powers of numbers, but rather descriptions of limits. The following examples, which were suggested by Prof. Jack Staib of Drexel University, show that such indeterminate forms can have any positive real value: (a) \(\lim _{x \rightarrow 0^{+}}\left[x^{(\ln a) /(1+\ln x)}\right]=a \quad\left(\right.\) form \(0^{0}\) ) (b) \(\lim _{x \rightarrow+\infty}\left[x^{(\ln a) /(1+\ln x)}\right]=a \quad\left(\right.\) form \(\infty^{0}\) ) (c) \(\lim _{x \rightarrow 0}\left[(x+1)^{(\ln a) / x}\right]=a \quad\left(\right.\) form \(\left.1^{\infty}\right)\). Verify these results.
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