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Chapter 2: Problem 12

Find any relative extrema of each function. List each extremum along with the \(x\) -value at which it occurs. Then sketch a graph of the function. $$ f(x)=3 x^{2}+2 x^{3} $$

Short Answer

Expert verified
Relative minimum at \(x = 0\) (value 0), and relative maximum at \(x = -1\) (value 1).

Step by step solution

01

Find the Derivative

To find the relative extrema, we first need to find the derivative of the function. The derivative of the function \(f(x) = 3x^2 + 2x^3\) is computed using the power rule. The power rule states that \(d/dx(x^n) = nx^{n-1}\). Applying the power rule:\[f'(x) = \frac{d}{dx}(3x^2) + \frac{d}{dx}(2x^3) = 6x + 6x^2\].
02

Find Critical Points

Critical points occur where the derivative equals zero or where the derivative is undefined. Since our derivative is a polynomial, it is defined everywhere. Set the derivative equal to zero and solve for \(x\):\[6x + 6x^2 = 0\]We can factor out \(6x\) to simplify:\[6x(x + 1) = 0\]Setting each factor to zero, we find the critical points: \(x = 0\) and \(x = -1\).
03

Determine the Nature of Critical Points

To determine whether each critical point is a relative minimum or maximum, we use the second derivative test. Find the second derivative:\[f''(x) = \frac{d}{dx}(6x + 6x^2) = 6 + 12x\]Evaluate the second derivative at each critical point:- For \(x = 0\):\[f''(0) = 6 + 12(0) = 6\] which is positive, indicating a relative minimum.- For \(x = -1\):\[f''(-1) = 6 + 12(-1) = -6\] which is negative, indicating a relative maximum.
04

Evaluate Function Values at Critical Points

To find the values of the function at the critical points, substitute \(x\) into the original function:- At \(x = 0\):\[f(0) = 3(0)^2 + 2(0)^3 = 0\]- At \(x = -1\):\[f(-1) = 3(-1)^2 + 2(-1)^3 = 3 - 2 = 1\]
05

Conclusion and Graph Sketch

The function has a relative minimum at \(x = 0\) with a function value of 0, and a relative maximum at \(x = -1\) with a function value of 1. To sketch the graph, plot these critical points and note their nature. The curve will descend to the minimum at \(x = 0\) and ascend to the maximum at \(x = -1\), forming a cubic curve typical of polynomials with terms \(x^2\) and \(x^3\).

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Key Concepts

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

Derivative
In calculus, the derivative of a function represents its rate of change or the slope of its tangent line at any given point. For polynomial functions, finding the derivative is straightforward thanks to the power rule. The power rule states that if you have a function of the form \(x^n\), its derivative is \(nx^{n-1}\). Applying this to the polynomial function \(f(x) = 3x^2 + 2x^3\), we find its derivative to be \(f'(x) = 6x + 6x^2\).

This derivative tells us how fast \(f(x)\) is changing at any point \(x\). Knowing the derivative is essential for locating critical points, where the function might change direction.

To summarize:
  • The derivative gives us the slope of the function at any point.
  • Using the power rule simplifies finding derivatives of polynomial terms.
Critical Points
Critical points occur where the derivative is zero or undefined. These are the points where the function's rate of change is zero, which means its graph has a horizontal tangent there.

In our exercise, the derivative \(f'(x) = 6x + 6x^2\) is a polynomial, meaning it is always defined. So we set it equal to zero to find the critical points:
\[6x + 6x^2 = 0\]Factor it like this:\[6x(x + 1) = 0\]From factoring, we find:
  • \(x = 0\)
  • \(x = -1\)
These are our critical points, places where the function potentially has a peak (maximum) or a trough (minimum).

Always inspect these points to understand the behavior of the function.
Second Derivative Test
The second derivative test helps determine whether a critical point is a relative minimum or maximum. We first find the second derivative by differentiating the first derivative function.

For our function:\[f''(x) = \frac{d}{dx}(6x + 6x^2) = 6 + 12x\] Evaluating this at each critical point provides insight into the nature of these points:
  • At \(x = 0\), \(f''(0) = 6\) (positive) indicates a relative minimum.
  • At \(x = -1\), \(f''(-1) = -6\) (negative) indicates a relative maximum.
A positive second derivative means the curve is concave up, showing a minimum, while a negative second derivative means it's concave down, showing a maximum.

Using the second derivative test simplifies classifying critical points.
Relative Extrema
Relative extrema refer to the relative minima and maxima of a function. These are the highest or lowest points in a small section of a graph. They differ from absolute extrema, which are the highest or lowest points over the entire domain.

In the given function, the critical points help us find the relative extrema:
  • The function has a relative minimum at \(x = 0\) with a value of \(f(0) = 0\).
  • It has a relative maximum at \(x = -1\) with a value of \(f(-1) = 1\).
To determine these points, we evaluated the function at each critical point found previously.

Understanding relative extrema is crucial for analyzing the behavior of functions on graphs.
Polynomial Functions
Polynomial functions are expressions built from variables and coefficients using only addition, subtraction, multiplication, and positive integer exponents. The given function, \(f(x) = 3x^2 + 2x^3\), is a polynomial function combining quadratic and cubic terms.

Key characteristics of polynomial functions include:
  • Smooth and continuous graphs without breaks or holes.
  • The degree of the polynomial determines the number of roots and critical points.
  • Derivatives of polynomial functions are also polynomials, simplifying analysis.
Polynomial functions like the one in our exercise form a fundamental part of calculus, making them important for describing real-world phenomena with smooth, predictable behaviors.

Always consider the degree and terms of polynomial functions for a deeper understanding of their graphs and potential extrema.

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Most popular questions from this chapter

Rate of change of the Arctic ice cap. In a trend that scientists attribute, at least in part, to global warming, the floating cap of sea ice on the Arctic Ocean has been shrinking since \(1980 .\) The ice cap always shrinks in summer and grows in winter. Average minimum size of the ice cap, in square miles, can be approximated by \(A=\pi r^{2}\) In \(2013,\) the radius of the ice cap was approximately \(792 \mathrm{mi}\) and was shrinking at a rate of approximately \(4.7 \mathrm{mi} / \mathrm{yr} .\) (Source: Based on data from nsidc.org.) How fast was the area changing at that time?

A power line is to be constructed from a power station at point \(A\) to an island at point \(C,\) which is \(l\) mi directly out in the water from \(a\) point \(B\) on the shore. Point \(B\) is 4 mi downshore from the power station at \(A\). It costs \(\$ 5000\) per mile to lay the power line under water and \(\$ 3000\) per mile to lay the line under ground. At what point \(S\) downshore from \(A\) should the line come to the shore in order to minimize cost? Note that \(S\) could very well be \(B\) or \(A\). (Hint: The length of \(C S\) is \(\left.\sqrt{1+x^{2}} .\right)\)

Find the absolute maximum and minimum values of each function, and sketch the graph. $$f(x)=\left\\{\begin{array}{ll}2 x+1 & \text { for }-3 \leq x \leq 1 \\\4-x^{2}, & \text { for } 1

Let $$y=(x-a)^{2}+(x-b)^{2}$$ For what value of \(x\) is \(y\) a minimum?

Find the absolute extrema of each function, if they exist, over the indicated interval. Also indicate the \(x\) -value at which each extremum occurs. When no interval is specified, use the real numbers, \((-\infty, \infty)\). $$f(x)=-0.01 x^{2}+1.4 x-30$$
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