/*! 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 38 Make a complete graph of the fol... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Make a complete graph of the following functions. A graphing utility is useful in locating intercepts, local extreme values, and inflection points. $$f(x)=\frac{1}{15} x^{3}-x+1$$

Short Answer

Expert verified
Answer: The y-intercept for the function is located at (0,1), and the inflection point is at x = 0.

Step by step solution

01

Determine the Domain and Range

Since the given function is a cubic polynomial function, its domain will be all real numbers. There are no restrictions on the input values for x. For the range, the function has no restrictions either, as the cubic term will allow the function to cover all possible y-values. Therefore, the domain and range of this function are both all real numbers.
02

Find the Intercepts

To find the x-intercepts, we have to set the function f(x) equal to zero and solve for x: $$0=\frac{1}{15} x^{3}-x+1$$ This is a cubic equation and might be difficult to solve algebraically. We could use a graphing utility to find the approximate x-intercepts. To find the y-intercept, we can set x to 0: $$f(0)=\frac{1}{15}(0)^3 - (0) + 1 = 1$$ Therefore, the y-intercept is (0,1).
03

Identify Local Extreme Values (Critical Points)

We need to find the critical points by taking the first derivative of the function and setting it equal to zero: $$f'(x) = \frac{d}{dx}(\frac{1}{15} x^{3} - x + 1) = \frac{1}{5}x^2 - 1$$ Now we set f'(x) to zero and solve for x: $$\frac{1}{5}x^2 - 1 = 0$$ This yields the solutions: $$x = \pm\sqrt{5}$$ These are the critical points of the function. To determine whether these points are local maxima or minima, we can use the second derivative test.
04

Determine Inflection Points

We need to take the second derivative of the function and set it equal to zero: $$f''(x) = \frac{d^2}{dx^2}(\frac{1}{15} x^{3} - x + 1) = \frac{2}{5}x$$ Now we set f''(x) to zero and solve for x: $$\frac{2}{5}x = 0$$ This yields the solution: $$x = 0$$ This is the inflection point of the function since the concavity of the graph changes at this point.
05

Make a Complete Graph

Now, we can put all this information together to create a complete graph of the given function: - The domain and range of the function are both all real numbers. - The y-intercept is located at (0,1). - The x-intercepts can be determined using a graphing utility. - There are two critical points at $$x=\pm\sqrt{5}$$, and we can use the second derivative test to identify whether these points correspond to local maxima or minima. - There is an inflection point at x = 0. Using all this information, one can make a complete graph of the function \(f(x)=\frac{1}{15} x^{3}-x+1\).

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!

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

Suppose you make a deposit of \(\$ P\) into a savings account that earns interest at a rate of \(100 \mathrm{r} \%\) per year. a. Show that if interest is compounded once per year, then the balance after \(t\) years is \(B(t)=P(1+r)^{t}\). b. If interest is compounded \(m\) times per year, then the balance after \(t\) years is \(B(t)=P(1+r / m)^{m t} .\) For example, \(m=12\) corresponds to monthly compounding, and the interest rate for each month is \(r / 12 .\) In the limit \(m \rightarrow \infty,\) the compounding is said to be continuous. Show that with continuous compounding, the balance after \(t\) years is \(B(t)=\overline{P e^{r t}}\).

The population of a species is given by the function \(P(t)=\frac{K t^{2}}{t^{2}+b},\) where \(t \geq 0\) is measured in years and \(K\) and \(b\) are positive real numbers. a. With \(K=300\) and \(b=30,\) what is \(\lim P(t),\) the carrying capacity of the population? b. With \(K=300\) and \(b=30,\) when does the maximum growth rate occur? c. For arbitrary positive values of \(K\) and \(b,\) when does the maximum growth rate occur (in terms of \(K\) and \(b\) )?

Determine the following indefinite integrals. Check your work by differentiation. $$\int\left(\csc ^{2} \theta+2 \theta^{2}-3 \theta\right) d \theta$$

Consider the quartic (fourth-degree) polynomial \(f(x)=x^{4}+b x^{2}+d\) consisting only of even-powered terms. a. Show that the graph of \(f\) is symmetric about the \(y\) -axis. b. Show that if \(b \geq 0\), then \(f\) has one critical point and no inflection points. c. Show that if \(b<0,\) then \(f\) has three critical points and two inflection points. Find the critical points and inflection points, and show that they alternate along the \(x\) -axis. Explain why one critical point is always \(x=0\) d. Prove that the number of distinct real roots of \(f\) depends on the values of the coefficients \(b\) and \(d,\) as shown in the figure. The curve that divides the plane is the parabola \(d=b^{2} / 4\) e. Find the number of real roots when \(b=0\) or \(d=0\) or \(d=b^{2} / 4\)

A mass oscillates up and down on the end of a spring. Find its position \(s\) relative to the equilibrium position if its acceleration is \(a(t)=\sin (\pi t),\) and its initial velocity and position are \(v(0)=3\) and \(s(0)=0,\) respectively.

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.