/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 If \(a>0, b>0,\) show that... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If \(a>0, b>0,\) show that \(f(x)=a\left(1-e^{-b x}\right)\) is everywhere increasing and everywhere concave down.

Short Answer

Expert verified
The function is everywhere increasing and everywhere concave down.

Step by step solution

01

Understand the function

The function is given as \(f(x) = a(1 - e^{-bx})\) with \(a > 0\) and \(b > 0\). Our goal is to show that this function is everywhere increasing and everywhere concave down.
02

Find the first derivative

The first derivative \(f'(x)\) of the function \(f(x) = a(1 - e^{-bx})\) is found using the chain rule. It is \(f'(x) = abe^{-bx}\).
03

Determine if the function is increasing

Since \(a > 0\) and \(b > 0\), we observe that \(f'(x) = ab e^{-bx}\) is always positive for all \(x\) because \(e^{-bx} > 0\) for all real numbers \(x\). Therefore, the function is everywhere increasing.
04

Find the second derivative

To check concavity, we find the second derivative \(f''(x)\). Differentiating \(f'(x) = abe^{-bx}\) again gives \(f''(x) = -ab^2 e^{-bx}\).
05

Determine concavity

The second derivative \(f''(x) = -ab^2 e^{-bx}\) is negative for all \(x\) since \(a > 0\), \(b > 0\), and \(e^{-bx} > 0\). This indicates that the function is everywhere concave down.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Derivatives and Their Importance
In calculus, the derivative of a function is a fundamental concept that measures how a function changes as its input changes. The derivative represents the rate of change or the slope of the function at any given point. For the function \(f(x) = a(1 - e^{-bx})\), the first derivative \(f'(x)\) is calculated using the chain rule, resulting in \(f'(x) = abe^{-bx}\).
This formula tells us the rate at which \(f(x)\) changes with respect to \(x\). Since exponential functions with negative exponents, like \(e^{-bx}\), always remain positive, it ensures that \(f'(x) > 0\) for all real \(x\) when \(a > 0\) and \(b > 0\).
  • The positivity of \(f'(x)\) indicates an increasing function.
  • Derivatives help in understanding various characteristics of functions, such as trends and rates of change.
Therefore, by analyzing \(f'(x)\), we can conclude that \(f(x)\) is always increasing.
Understanding Concavity in Functions
Concavity describes the direction in which a function curves. A function is concave down if its graph opens downwards like a frown, and concave up if it opens upwards like a smile. To determine concavity, the second derivative \(f''(x)\) is examined. For our function, \(f''(x) = -ab^2 e^{-bx}\).
The sign of \(f''(x)\) provides insight into the concavity:
  • If \(f''(x) < 0\), the function is concave down.
  • Conversely, if \(f''(x) > 0\), the function is concave up.
In this case, since \(a > 0\) and \(b > 0\), the component \(-ab^2 e^{-bx}\) is inherently negative for all real \(x\).
Thus, it is established that the function \(f(x)\) does not change its concavity and is always concave down.
Properties of Increasing Functions
An increasing function is one where the value of the function grows as the input grows. It implies that as \(x\) moves to larger values, \(f(x)\) also moves to larger values. For a function to be classified as increasing over its entire domain, its first derivative must be non-negative for all values of \(x\).
For \(f(x) = a(1 - e^{-bx})\), the first derivative \(f'(x) = ab e^{-bx}\) assures us that it is positive, provided \(a > 0\) and \(b > 0\).
  • A positive derivative means the function climbs upward as \(x\) increases.
  • This growing trend is uniform for every point on the graph.
In summary, exploring the first derivative reveals \(f(x)\) to be increasing across the entire domain of \(x\).

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find formulas for the functions described. A function of the form \(y=a\left(1-e^{-b x}\right)\) with \(a, b>0\) and a horizontal asymptote of \(y=5\).

The Lennard-Jones model predicts the potential energy \(V(r)\) of a two-atom molecule as a function of the distance \(r\) between the atoms to be $$V(r)=\frac{A}{r^{12}}-\frac{B}{r^{6}}, \quad r>0$$ where \(A\) and \(B\) are positive constants. (a) Evaluate \(\lim _{r \rightarrow 0^{+}} V(r),\) and interpret your answer. (b) Find the critical point of \(V(r) .\) Is it a local maximum or local minimum? (c) The inter-atomic force is given by \(F(r)=-V^{\prime}(r)\) At what distance \(r\) is the inter-atomic force zero? (This is called the equilibrium size of the molecule.) (d) Describe how the parameters \(A\) and \(B\) affect the equilibrium size of the molecule.

Explain what is wrong with the statement. An increasing function has no inflection points.

Are the statements in Problems true or false for a function \(f\) whose domain is all real numbers? If a statement is true,explain how you know. If a statement is false, give a counterexample. If \(x=p\) is not a local maximum of \(f,\) then \(x=p\) is not a critical point of \(f\).

Investigate the given two parameter family of functions. Assume that \(a\) and \(b\) are positive. (a) Graph \(f(x)\) using \(b=1\) and three different values for \(a\). (b) Graph \(f(x)\) using \(a=1\) and three different values for \(b\). (c) In the graphs in parts (a) and (b), how do the critical points of \(f\) appear to move as \(a\) increases? As \(b\) increases? (d) Find a formula for the \(x\) -coordinates of the critical point(s) of \(f\) in terms of \(a\) and \(b\). $$f(x)=(x-a)^{2}+b$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.