Problem 16
Question
Suppose that \(f(a)=g(a)\) and that \(f^{\prime}(x) \leq g^{\prime}(x)\) for all \(x \geq a\). (a) Prove that \(f(x) \leq g(x)\) for all \(x \geq a\). (b) Use part (a) to prove that \(e^{x} \geq 1+x\) for all \(x \geq 0\). (c) Use parts (a) and (b) to prove that \(e^{x} \geq 1+x+\frac{x^{2}}{2}\) for all \(x \geq 0\). (d) Can you generalize these results?
Step-by-Step Solution
Verified Answer
(a) \( f(x) \leq g(x) \). (b) \( e^x \geq 1+x \). (c) \( e^x \geq 1+x+\frac{x^2}{2} \). (d) Yes, through Taylor series.
1Step 1: Understanding the Given Information
We have two functions, \( f(x) \) and \( g(x) \), such that \( f(a) = g(a) \) and \( f^{\prime}(x) \leq g^{\prime}(x) \) for all \( x \geq a \). Our goal is to show \( f(x) \leq g(x) \) for all \( x \geq a \).
2Step 2: Use of Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus connects integrals and derivatives. Consider the derivative condition \( f^{\prime}(x) \leq g^{\prime}(x) \). Integrate both sides from \( a \) to \( x \):\[ \int_{a}^{x} f^{\prime}(t) \, dt \leq \int_{a}^{x} g^{\prime}(t) \, dt. \] This implies \( f(x) - f(a) \leq g(x) - g(a) \). Since \( f(a) = g(a) \), it follows that \( f(x) \leq g(x) \).
3Step 3: Proving Exponential Inequality
To use part (a), define \( f(x) = e^x \) and \( g(x) = 1 + x \). At \( x = 0 \), \( f(0) = g(0) = 1 \). Also, \( f^{\prime}(x) = e^x \geq 1 = g^{\prime}(x) \). Using the previous result, \( e^x \geq 1 + x \) for all \( x \geq 0 \).
4Step 4: Extending to Quadratic Polynomial
Next, define \( f(x) = e^x \) and \( g(x) = 1 + x + \frac{x^2}{2} \). At \( x = 0 \), \( f(0) = 1 \) and \( g(0) = 1 \). Calculating derivatives, \( f^{\prime}(x) = e^x \) and \( g^{\prime}(x) = 1 + x \). Since \( e^x \geq 1 + x \) (from Step 3), the condition for part (a) is satisfied, and therefore \( e^x \geq 1 + x + \frac{x^2}{2} \) for all \( x \geq 0 \).
5Step 5: Generalizing the Pattern
Notice that we can extend this pattern by using Taylor series for \( e^x \). We have shown \( e^x \geq 1+x \) and \( e^x \geq 1+x+\frac{x^2}{2} \). Generally, \( e^x \geq 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} \) for all \( x \geq 0 \), where each term is derived from the Taylor expansion of \( e^x \).
Key Concepts
Fundamental Theorem of CalculusExponential FunctionsTaylor SeriesInequality Proofs
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a crucial part of calculus that connects differentiation with integration. It sets up a powerful method for evaluating definite integrals as well as understanding the relationship between derivatives and integrals. Consider two functions, \( f(x) \) and \( g(x) \), such that \( f'(x) \leq g'(x) \) for \( x \geq a \). This tells us that the rate of change of \( f(x) \) is always less or equal to that of \( g(x) \).
By integrating both sides from \( a \) to \( x \), we align with the theorem which says:
By integrating both sides from \( a \) to \( x \), we align with the theorem which says:
- \( \int_a^x f'(t) \, dt \leq \int_a^x g'(t) \, dt \).
- Upon integration, this results in \( f(x) - f(a) \leq g(x) - g(a) \).
Exponential Functions
Exponential functions are immensely significant in mathematics. A classic example is \( e^x \), where \( e \) is the base of the natural logarithm. It represents continuous growth. We particularly study these functions not just for their algebraic simplicity but also for their behavior in growth models.
To show \( e^x \geq 1 + x \) for \( x \geq 0 \), note the properties and behaviors of the function \( e^x \).
To show \( e^x \geq 1 + x \) for \( x \geq 0 \), note the properties and behaviors of the function \( e^x \).
- At \( x = 0 \), both \( e^x \) and \( 1 + x \) evaluate to 1, setting the comparison's base.
- The derivative, \( f'(x) = e^x \), characterizes the slope of \( e^x \).
- It is evident that \( e^x \) grows faster than or equal to any linear addition starting from any positive x-axis point.
Taylor Series
The Taylor series is a beautiful tool allowing functions to be expressed as infinite series expansions. The series representation of \( e^x \) converges remarkably well for all real \( x \). Consider the expansion:
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \] This expansion proves useful in approximations and proof of inequalities, like in the case of \( e^x \geq 1 + x + \frac{x^2}{2} \).
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \] This expansion proves useful in approximations and proof of inequalities, like in the case of \( e^x \geq 1 + x + \frac{x^2}{2} \).
- The initial terms give an approximation that adheres closely to the function's graph.
- By analyzing each successive term, the Taylor series can become arbitrarily close to the actual function value.
- Each term \( \frac{x^n}{n!} \) has a decreasing influence on the sum's total, which is why early terms set accurate bounds.
Inequality Proofs
Inequality proofs are foundational in mathematics, providing methods to compare quantities or mathematical expressions. These proofs rely on logically deduced steps to arrive at certain bounds or restrict functions in given domains. When dealing with functions like \( f(x) = e^x \) and \( g(x) = 1 + x \) or \( g(x) = 1 + x + \frac{x^2}{2} \), our process is systematic.
Begin by ensuring initial conditions fit the equation setup, for instance, starting at \( x = 0 \) for comparative simplicity where needed.
Begin by ensuring initial conditions fit the equation setup, for instance, starting at \( x = 0 \) for comparative simplicity where needed.
- Assess whether function derivatives support the inequality throughout the desired interval.
- Utilize the outcomes of proven inequalities—such as \( f'(x) \leq g'(x) \)—as stepping stones for extended proofs.
Other exercises in this chapter
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