Chapter 4: Problem 84
a. Prove that \(e^{x} \geq 1+x\) if \(x \geq 0\) b. Use the result in part (a) to show that $$e^{x} \geq 1+x+\frac{1}{2} x^{2}$$
Short Answer
Expert verified
(a) Use the derivative to show \(f(x) = e^x - (1+x) \geq 0\) for \(x \geq 0\). (b) Extend to show \(g(x) = e^x - (1+x+\frac{1}{2}x^2) \geq 0\) using part (a).
Step by step solution
01
Understanding the Base Case
For this inequality proof, start by considering the function \(f(x) = e^x - (1+x)\). We aim to show \(f(x) \geq 0\) for \(x \geq 0\). Initially, check at \(x = 0\). Here, \(f(0) = e^0 - (1+0) = 1 - 1 = 0\), so the base case holds.
02
Applying Differentiation to Examine Behavior
To demonstrate that \(f(x)\) is non-negative for \(x \geq 0\), analyze its derivative. Calculate \(f'(x) = \frac{d}{dx}(e^x - (1+x)) = e^x - 1\). For \(x \geq 0\), notice that \(e^x \geq 1\), thus \(f'(x) \geq 0\). This implies that \(f(x)\) is non-decreasing.
03
Analyzing Function Behavior for Conclusion
Since \(f(x)\) is non-decreasing and \(f(0) = 0\), it follows that for \(x \geq 0\), \(f(x) = e^x - (1+x) \geq 0\). Therefore, \(e^x \geq 1+x\) is proven for \(x \geq 0\).
04
Extending the Result to a Stronger Inequality
Consider \(g(x) = e^x - (1+x+\frac{1}{2}x^2)\). We now need to demonstrate \(g(x) \geq 0\) for \(x \geq 0\). Calculate its derivative \(g'(x) = e^x - (1+x)\). From previous steps, \(g'(x) = f(x) \geq 0\) for \(x \geq 0\). Thus, \(g(x)\) is non-decreasing.
05
Verifying Initial Condition for New Function
Check \(g(x)\) at \(x = 0\): \(g(0) = e^0 - (1+0+\frac{1}{2}(0)^2) = 1 - 1 = 0\). Since \(g(x)\) is non-decreasing and \(g(0) = 0\), then \(g(x) \geq 0\) for \(x \geq 0\).
06
Concluding the Extended Inequality
Thus, \(e^x \geq 1+x+\frac{1}{2}x^2\) is valid for \(x \geq 0\), using the results from part (a) and the fact that \(g(x)\) is non-decreasing from 0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Functions
Exponential functions are fundamental in mathematics and appear in numerous real-world scenarios, such as continuous growth or decay processes like population studies or radioactive decay. An exponential function is typically expressed in the form \(y = a^x\), where \(a\) is a constant and \(x\) is a variable. The most common base for these functions is the number \(e\), approximately equal to 2.718, which serves as the foundation for natural exponential functions.
The function \(e^x\) grows exponentially as \(x\) increases, meaning its rate of increase escalates rapidly. This characteristic makes \(e^x\) a powerful mathematical tool, particularly in calculus, to model continuous compounding processes.
Key features include:
The function \(e^x\) grows exponentially as \(x\) increases, meaning its rate of increase escalates rapidly. This characteristic makes \(e^x\) a powerful mathematical tool, particularly in calculus, to model continuous compounding processes.
Key features include:
- The function \(e^x\) is positive for all real \(x\).
- It passes through the point (0,1), meaning \(e^0 = 1\).
- \(e^x\) is its own derivative, which means the rate of change of \(e^x\) is equal to itself.
Derivative Analysis
Derivative analysis is a fundamental concept in calculus that helps us understand how functions change. It involves finding the derivative, which represents the rate at which a function's value changes with respect to its input. For challenging scenarios like inequalities, derivatives are instrumental. To analyze a function's behavior, we often differentiate it to see how it changes across intervals.
In the given exercise, we consider the function \(f(x) = e^x - (1+x)\) to prove an inequality. By differentiating it, we get \(f'(x) = e^x - 1\). The derivative \(f'(x)\) tells us whether \(f(x)\) is increasing or decreasing. For \(x \geq 0\), \(f'(x) = e^x - 1\) is always non-negative since \(e^x \geq 1\). This shows that \(f(x)\) never decreases in this range.
Key points to remember about derivative analysis:
In the given exercise, we consider the function \(f(x) = e^x - (1+x)\) to prove an inequality. By differentiating it, we get \(f'(x) = e^x - 1\). The derivative \(f'(x)\) tells us whether \(f(x)\) is increasing or decreasing. For \(x \geq 0\), \(f'(x) = e^x - 1\) is always non-negative since \(e^x \geq 1\). This shows that \(f(x)\) never decreases in this range.
Key points to remember about derivative analysis:
- If the derivative of a function is positive, the function is increasing.
- If the derivative is zero, the function is constant over that interval.
- If the derivative is negative, the function is decreasing.
Inequality Proofs
Inequality proofs often involve showing that one expression is greater than or equal to another for a certain range of values, which requires a combination of calculus and algebraic techniques. The goal is to demonstrate a general truth across an interval rather than just at specific points.
In the context of the exercise, we're asked to prove that \(e^x \geq 1+x\) for \(x \geq 0\) in part (a), and then extend it in part (b) to \(e^x \geq 1+x+\frac{1}{2}x^2\). To achieve this, we define new functions and analyze their behavior through derivative analysis.
Important considerations for inequality proofs include:
In the context of the exercise, we're asked to prove that \(e^x \geq 1+x\) for \(x \geq 0\) in part (a), and then extend it in part (b) to \(e^x \geq 1+x+\frac{1}{2}x^2\). To achieve this, we define new functions and analyze their behavior through derivative analysis.
Important considerations for inequality proofs include:
- Setting up functions that capture the essence of the inequality. For instance, \(f(x) = e^x - (1+x)\).
- Checking boundary or initial conditions, such as \(x = 0\), to establish a base case.
- Using differentiation to ascertain if the function is non-decreasing or non-increasing to extend the observed behavior across a specified range.