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Chapter 5: Problem 21

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 \(x=c\).

Short Answer

Expert verified
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

01

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.
02

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, \( g'(c) = e^{f(c)} \, f'(c) = e^{f(c)} \, 0 = 0 \).
03

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 \( g''(c) = e^{f(c)} \, 0 + e^{f(c)} \, f''(c) = e^{f(c)} \, f''(c) \).
04

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 \( x = c \).

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Key Concepts

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

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 鈥� 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.
Understanding differentiable functions is crucial for analyzing the behavior of a function, especially when examining critical points like local maxima and minima.
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.
Recognizing and determining these critical points help in understanding the function's graph and behavior.
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.
Mastering the chain rule helps in differentiating intricate expressions, unlocking deeper insights into the function's behavior.

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Most popular questions from this chapter

Find the general antiderivative of the given function. $$ f(x)=-3 \sin \left(\frac{\pi}{3} x\right)+4 \cos \left(-\frac{\pi}{4} x\right) $$

Fish are indeterminate growers; that is, their length \(L(t)\) increases with age \(t\) throughout their lifetime. If we plot the growth rate \(d L / d t\) versus age \(t\) on semilog paper, a straight line with negative slope results, meaning that: $$ \frac{d L}{d t}=A e^{-k t} $$ where \(A>0\) and \(k>0\) are both coefficients that depend on the species of fish, and the habitat that it is growing in. (a) Find the solution for this differential equation (your solution will include \(A\) and \(k\) as unknown constants, as well as one additional unknown constant \(C\) from the antiderivative). (b) Find the values for the constants \(A, k, C\), that would fit the solution to the following data \(L(0)=5, L(1)=10\), and $$ \lim _{t \rightarrow \infty} L(t)=30. $$ (c) Graph the solution \(L(t)\) as a function of \(t\).

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Suppose that a drug is eliminated so slowly from the blood that its elimination kinetics can be essentially ignored. Then according to Section \(5.9\) the total amount of drug in the blood is given by a differential equation: $$ \frac{d M}{d t}=A(t) $$ where \(A(t)\) is the rate of absorption. We will show in Chapter 8 that if the drug is absorbed into the blood from a pill in the patient's gut, then \(A(t)\) is given by a function $$ A(t)=C e^{-k t} $$ where \(C>0\) and \(k>0\) are constants that depend on the type of the drug being administered. Assume that at \(t=0\) there is no drug present in the patient's blood (i.e., \(M(0)=0\) ). Solve this initial value problem, and, using the methods from Section \(5.6\), sketch the graph of \(M(t)\) against \(t\).

Find the general antiderivative of the given function. $$ f(x)=\cos ^{2} x-\sin ^{2} x $$
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