Problem 21
Question
Suppose that \(f(x)\) is differentiable on \(\mathbf{R}\). Show that if \(f(x)\) has a local maximum at \(x=c\), then \(g(x)=e^{f(x)}\) also has a local maximum at \(x=c\).
Step-by-Step Solution
Verified Answer
If \( f(x) \) has a local maximum at \( x = c \), then \( g(x) = e^{f(x)} \) also has a local maximum at \( x = c \).
1Step 1: Understand the Concepts
To solve this problem, we need to use the properties of differentiable functions and what it means for a function to have a local maximum. Specifically, if a function \( f(x) \) has a local maximum at \( x = c \), the derivative \( f'(x) \) at \( x=c \) should be zero. We will examine how this impacts the behavior of the function \( g(x) = e^{f(x)} \) at the same point.
2Step 2: Analyze the Derivative of g(x)
To find if \( g(x) = e^{f(x)} \) has a local maximum at \( x = c \), we should take the derivative \( g'(x) \). By the chain rule, \( g'(x) = e^{f(x)} \, f'(x) \). Since \( e^{f(x)} > 0 \) for all real numbers, the sign of \( g'(x) \) depends on \( f'(x) \). Since there is a local maximum of \( f(x) \) at \( x = c \), \( f'(c) = 0 \). Thus, \( g'(c) = e^{f(c)} \, f'(c) = e^{f(c)} \, 0 = 0 \).
3Step 3: Investigate the Second Derivative
To further analyze if \( g(x) \) has a local maximum, we look at the second derivative \( g''(x) \). First, we derive \( g'(x) = e^{f(x)} \, f'(x) \). Using the product rule, \( g''(x) = e^{f(x)} \, (f'(x))^2 + e^{f(x)} \, f''(x) \). At \( x = c \), \( f'(c) = 0 \), so \( g''(c) = e^{f(c)} \, 0 + e^{f(c)} \, f''(c) = e^{f(c)} \, f''(c) \).
4Step 4: Conclusion from the Sign of the Second Derivative
Since \( f(x) \) has a local maximum at \( x = c \), \( f''(c) \leq 0 \). Hence, \( g''(c) = e^{f(c)} \, f''(c) \leq 0 \). The positive factor \( e^{f(c)} \) does not affect the sign of \( g''(c) \). Thus, if \( f(x) \) has a local maximum at \( x = c \), \( g(x) = e^{f(x)} \) will also have a local maximum at \( x = c \).
Key Concepts
Differentiable FunctionsLocal MaximumChain Rule
Differentiable Functions
The concept of a differentiable function is fundamental in calculus. A function is considered differentiable at a point if it has a derivative there. This basically means that we can draw a tangent line at that specific point on the function's curve. Differentiability implies that the function is smooth and continuous around the point, without any breaks or sharp corners.
- If a function is differentiable, it is also continuous. However, the reverse is not always true — a continuous function is not necessarily differentiable.
- The derivative of a function at any point gives us the slope of the tangent line to the curve of the function at that point.
Local Maximum
A local maximum of a function occurs at a point where the function value is higher than at any other point in its immediate neighborhood. Let's break this down further:
- At a local maximum, moving slightly to the left or right causes the function value to decrease.
- Mathematically, if a function has a local maximum at point \(x = c\), then the derivative \(f'(c) = 0\). This is because the slope of the function's tangent at that point becomes horizontal.
- Additionally, around a point of local maximum, the second derivative, \(f''(x)\), must be less than or equal to zero. This reveals information about the concavity of the function, indicating it curves downward.
Chain Rule
The chain rule is a powerful tool in calculus used to differentiate composite functions. When a function is composed of two others, knowing how to differentiate each can allow us to tackle the more complex form.
- The chain rule states: If \(y = f(g(x))\), then the derivative \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\).
- In our problem, where \(g(x) = e^{f(x)}\), the chain rule helps differentiate \(g(x)\) by utilizing \(f(x)\) and its derivative.
- This application of the chain rule allows us to consider both the external function \(e^x\) and the internal function \(f(x)\) simultaneously.
Other exercises in this chapter
Problem 20
Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow 0} \frac{e^{x}-1}{3 e^{2 x}-x} $$
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Find the general antiderivative of the given function. $$ f(x)=2 e^{2 x} $$
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Sketch the graph of (a) a function that is increasing at an accelerating rate; and (b) a function that is increasing at a decelerating rate. (c) Assume that you
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In Problems , find c such that \(f^{\prime}(c)=0\) and determine whether \(f(x)\) has a local extremum at \(x=c .\) $$ f(x)=-x^{2} $$
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