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Find the exact location of all the relative and absolute extrema of each function. \(f(t)=-2 t^{3}-3 t\) with domain \([-1,1]\)

Short Answer

Expert verified
The short answer is: The function \(f(t)=-2t^3-3t\) has a critical point at \(t=0\) within the domain \([-1,1]\). The second derivative \(f''(t)=-12t\) is positive when \(t=0\), so this is a relative minimum. The values of the function at the endpoints are \(f(-1)=1\) and \(f(1)=-1\). Since \(f(0)=0\) and lies between the endpoint values, the absolute minimum is \(f(0)=0\) and the absolute maximum occurs at \(f(-1)=1\).

Step by step solution

01

Find the first derivative of the function

To find the extrema of the function, first, find its first derivative with respect to t: \(f(t) = -2t^3 - 3t\) The derivative of f(t) with respect to t is: \(f'(t) = \frac{d}{dt}(-2t^3 - 3t)\)
02

Find the critical points by solving the equation f'(t) = 0

Now we need to find the points where the derivative is equal to 0, as those are the critical points where the function might have extrema. To do this, we set f'(t) to 0 and solve for t: \(0 = -6t^2 - 3\)
03

Evaluate the second derivative to check for relative extrema

To find whether a critical point t is a relative maximum, minimum, or neither, we can evaluate the second derivative of the function, following this rule: If the second derivative is positive at the critical point, it is a relative minimum; if negative, it is a relative maximum; if zero, it's inconclusive. The second derivative of f(t) with respect to t is: \(f''(t) = \frac{d^2}{dt^2}(-2t^3 - 3t)\)
04

Determine absolute extrema by comparing the values of f(t) at critical points and domain endpoints

To find the absolute extrema, test whether the critical points fall within the given domain by comparing the values of f(t) at the critical points and the domain endpoints. If the critical points lie within the domain, then the absolute extrema will occur at either the endpoints or the critical points. Domain: \([-1, 1]\) Evaluate f(t) at the critical points and endpoints: \(f(-1) = \) \(f(1) = \) Compare the above values to determine the absolute extrema of the function.

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Key Concepts

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

Critical Points
Critical points are crucial for analyzing the behavior of functions. In the context of calculus, they are where the function's first derivative is either zero or undefined. These points can be potential locations for relative extrema, which are the high or low points on a function within a given interval.

For the given function f(t) = -2t^3 - 3t, we found its first derivative f'(t) = -6t^2 - 3. To find the critical points, we set this first derivative equal to zero and solved the equation 0 = -6t^2 - 3. The solution to this equation gives us the values of t at which the function's slope is zero and thus where possible extrema may occur. It is important to ensure these points are within the given domain. Critical points are a key first step not only in finding extremes but also in understanding the overall shape and turning points of the function.
First Derivative Test
The First Derivative Test is a powerful tool used to determine whether a critical point is a relative maximum or minimum. The test involves looking at the sign (positive or negative) of the first derivative before and after the critical point. If the derivative changes from positive to negative at the critical point, the function has a relative maximum there. Conversely, if it changes from negative to positive, that point is a relative minimum.

In our exercise, after finding the critical points, we would apply the First Derivative Test to each one by checking the sign of f'(t) on each side of the critical points. Points that are not critical points can be skipped in this test; we focus solely on those that could be potential extrema. While the First Derivative Test can identify relative extrema, it does not provide information on whether these are absolute extrema -- the highest or lowest points over the entire domain.
Second Derivative Test
The Second Derivative Test is another method for determining whether a critical point is a relative maximum, minimum, or a point of inflection. This test involves the second derivative of the function, f''(t). If f''(t) is greater than zero at a critical point, the function is concave up at that point, indicating a relative minimum. If f''(t) is less than zero, the function is concave down, suggesting a relative maximum. If f''(t) is zero, the test is inconclusive, and one may need to use other methods to determine the nature of the critical point.

In the provided exercise, we calculated the second derivative and used it to evaluate the nature of the critical points. This helps us to not just pinpoint the location of the potential extrema but also to classify them properly. Remember that while relative extrema are local to the neighborhood around critical points, absolute extrema consider the entire domain of the function and are identified by comparing values at critical points and at the endpoints of the domain.

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Most popular questions from this chapter

A general quadratic demand function has the form \(q=a p^{2}+b p+c(a, b\), and \(c\) constants with \(a \neq 0\) ). a. Obtain a formula for the price elasticity of demand at a unit price \(p\). b. Obtain a formula for the price or prices that could maximize revenue.

Company C's profits satisfy \(P(0)=\$ 1\) million, \(P^{\prime}(0)=\) \(\$ 1\) million per year, and \(P^{\prime \prime}(0)=-\$ 1\) million per year per year. Company D's profits satisfy \(P(0)=\$ 0\) million, \(P^{\prime}(0)=\) \(\$ 0\) million per year, and \(P^{\prime \prime}(0)=\$ 1\) million per year per year. There are no points of inflection in either company's profit curve. Sketch two pairs of profit curves: one in which Company C ultimately outperforms Company \(\mathrm{D}\) and another in which Company D ultimately outperforms Company C.

A time- series study of the demand for higher education, using tuition charges as a price variable, yields the following result: $$\frac{d q}{d p} \cdot \frac{p}{q}=-0.4$$ where \(p\) is tuition and \(q\) is the quantity of higher education. \(\mathbf{2 5}\). Which of the following is suggested by the result? (A) As tuition rises, students want to buy a greater quantity \(\quad \mathbf{2 6}\). of education. (B) As a determinant of the demand for higher education, income is more important than price. (C) If colleges lowered tuition slightly, their total tuition receipts would increase. (D) If colleges raised tuition slightly, their total tuition receipts would increase. (E) Colleges cannot increase enrollments by offering larger scholarships.

You have been hired as a marketing consultant to Big Book Publishing, Inc., and you have been approached to determine the best selling price for the hit calculus text by Whiner and Istanbul entitled Fun with Derivatives. You decide to make life easy and assume that the demand equation for Fun with Derivatives has the linear form \(q=m p+b\), where \(p\) is the price per book, \(q\) is the demand in annual sales, and \(m\) and \(b\) are certain constants you'll have to figure out. a. Your market studies reveal the following sales figures: when the price is set at \(\$ 50.00\) per book, the sales amount to 10,000 per year; when the price is set at \(\$ 80.00\) per book, the sales drop to 1000 per year. Use these data to calculate the demand equation. b. Now estimate the unit price that maximizes annual revenue and predict what Big Book Publishing, Inc.'s annual revenue will be at that price.

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