Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Connecting \(f\) and \(\boldsymbol{f}^{\prime}\) Sketch a smooth curve \(y=f(x)\) through the origin with the properties that \(f^{\prime}(x)<0\) for \(x<0\) and \(f^{\prime}(x)>0\) for \(x>0\)

Short Answer

Expert verified
The curve representing \(f(x)\) passes through the origin and decreases for \(x<0\) while increasing for \(x>0\). This shows the relationship between the function and its derivative, in that the negative slope corresponds to a negative rate of change (or derivative), and a positive slope corresponds to a positive rate of change.

Step by step solution

01

Understand the Properties of Derivatives

Given that the derivative of a function represents the function's rate of change, when the derivative of a function is negative for \(x<0\), this means that for all negative \(x\) values, the function is decreasing. Similarly, if the derivative of a function is positive for \(x>0\), this means the function is increasing for all positive \(x\) values.
02

Sketch the Curve

Based on these properties, sketch a curve that passes through the origin. The curve should decrease for \(x<0\), indicating a negative slope or rate of change, and increase for \(x>0\), indicating a positive slope or rate of change. This curve will represent the function \(y=f(x)\).
03

Interpret the Curve

The resulting curve should illustrate the properties of the derivatives described in the problem. For \(x<0\), the curve decreases, showing the rate of change is negative. For \(x>0\), the curve increases, showing the rate of change is positive. This graphical representation evidences the connection between \(f(x)\) and \(f^{\prime}(x)\).

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 Vaia!

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

True or False If \(f^{\prime}(c)=0\) and \(f^{\prime \prime}(c)<0,\) then \(f(c)\) is a local maximum. Justify your answer.

\(f\) is an even function, continuous on \([-3,3],\) and satisfies the following. (d) What can you conclude about \(f(3)\) and \(f(-3) ?\)

The Linearization is the Best Linear Approximation Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=m(x-a)+c(m\) and \(c\) constants). If the error \(E(x)=f(x)-g(x)\) were small enough near \(x=a,\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions i. \(E(a)=0\) ii. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \((x-a)\) .

Multiple Choice A cylindrical rubber cord is stretched at a constant rate of 2 \(\mathrm{cm}\) per second. Assuming its volume does no change, how fast is its radius shrinking when its length is 100 \(\mathrm{c}\) and its radius is 1 \(\mathrm{cm} ?\) $$\begin{array}{ll}{\text { (A) } 0 \mathrm{cm} / \mathrm{sec}} & {\text { (B) } 0.01 \mathrm{cm} / \mathrm{sec}} 67 {\text{ (C) } 0.02 \mathrm{cm} / \mathrm{sec}}$\\\ {\text { (D) } 2 \mathrm{cm} / \mathrm{sec}} & {\text { (E) } 3.979 \mathrm{cm} / \mathrm{sec}}\end{array}

Multiple Choice The \(x\) -coordinates of the points of inflection of the graph of \(y=x^{5}-5 x^{4}+3 x+7\) are \(\mathrm \) (A) 0 only (B) 1 only (C) 3 only (D) 0 and 3 (E) 0 and 1

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.

Sign-up for free