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Chapter 5: Problem 3

(a) Use both the first and second derivative tests to show that \(f(x)=3 x^{2}-6 x+1\) has a relative minimum at \(x=1\) (b) Use both the first and second derivative tests to show that \(f(x)=x^{3}-3 x+3\) has a relative minimum at \(x=1\) and a relative maximum at \(x=-1\)

Short Answer

Expert verified
(a) \(x=1\) is a relative minimum for \(f(x)=3x^2-6x+1\). (b) \(x=1\) is a relative minimum and \(x=-1\) is a relative maximum for \(f(x)=x^3-3x+3\).

Step by step solution

01

Determine the first derivative of \(f(x)=3x^{2}-6x+1\)

Calculate the first derivative \(f'(x)=\frac{d}{dx}(3x^{2}-6x+1)\). The result is:\[ f'(x) = 6x - 6 \]
02

Solve \(f'(x)=0\) for critical points

Set \(f'(x)=6x-6=0\) and solve for \(x\). This gives \(6x=6\), so \(x=1\) is a critical point.
03

Apply the First Derivative Test for \(f(x)=3x^{2}-6x+1\)

Evaluate the sign of \(f'(x)\) around \(x=1\). For \(x<1\), choose \(x=0\): \(f'(0) = 6(0) - 6 = -6\) (negative). For \(x>1\), choose \(x=2\): \(f'(2) = 6(2) - 6 = 6\) (positive). The function changes from decreasing (-) to increasing (+), indicating a relative minimum at \(x=1\).
04

Determine the second derivative of \(f(x)=3x^{2}-6x+1\)

Calculate the second derivative \(f''(x)=\frac{d^2}{dx^2}(3x^{2}-6x+1)\). The result is:\[ f''(x) = 6 \]
05

Apply the Second Derivative Test for \(f(x)=3x^{2}-6x+1\)

Evaluate \(f''(x)\) at \(x=1\): \(f''(1) = 6\). Since \(f''(1) > 0\), \(f(x)\) is concave up at \(x=1\), confirming a relative minimum.
06

Determine the first derivative of \(f(x)=x^{3}-3x+3\)

Calculate the first derivative \(f'(x)=\frac{d}{dx}(x^{3}-3x+3)\). The result is:\[ f'(x) = 3x^{2} - 3 \]
07

Solve \(f'(x)=0\) for critical points

Set \(f'(x)=3x^{2} - 3 = 0\) and solve for \(x\). This gives \(3(x^{2}-1)=0\), leading to \(x^{2}=1\), so \(x=\pm1\) are critical points.
08

Apply the First Derivative Test for \(f(x)=x^{3}-3x+3\)

Evaluate the sign of \(f'(x)\) around the critical points. For \(x=-2\), \(f'(-2) = 9\) (positive); for \(x=0\), \(f'(0) = -3\) (negative); for \(x=2\), \(f'(2) = 9\) (positive). The function changes from increasing to decreasing at \(x=-1\), indicating a relative maximum, and from decreasing to increasing at \(x=1\), indicating a relative minimum.
09

Determine the second derivative of \(f(x)=x^{3}-3x+3\)

Calculate the second derivative \(f''(x)=\frac{d^2}{dx^2}(x^{3}-3x+3)\). The result is:\[ f''(x) = 6x \]
10

Apply the Second Derivative Test for \(f(x)=x^{3}-3x+3\) at each critical point

Evaluate \(f''(x)\) at the critical points. For \(x=-1\), \(f''(-1) = -6\) (negative), confirming a relative maximum. For \(x=1\), \(f''(1) = 6\) (positive), confirming a relative minimum.

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Key Concepts

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

First Derivative Test
The first derivative test is a helpful way to identify local maxima and minima by examining how a function's slope changes at or around critical points. The steps are simple:
  • First, find the first derivative of the function. This tells us the rate at which the function is changing.
  • Next, solve for critical points by setting the first derivative equal to zero. Critical points occur where the slope of the tangent line to the curve is zero, indicating a potential max or min.
  • Finally, check the sign of the first derivative on either side of each critical point. If it changes from negative to positive, there's a relative minimum. If it changes from positive to negative, there's a relative maximum.
For example, in the function \(f(x)=3x^{2}-6x+1\), the first derivative is \(6x-6\). Setting it equal to zero gives \(x=1\) as a critical point. Checking points around \(x=1\), such as \(x=0\) and \(x=2\), we find the derivative changes sign from negative to positive, confirming a relative minimum at \(x=1\). This change in sign is the hallmark of the first derivative test.
Second Derivative Test
The second derivative test provides an alternative method for finding local maxima and minima by indicating the concavity of a function at its critical points.
  • Start by finding the second derivative of the function. The second derivative gives us a sense of the rate of change of the slope, or in simpler terms, how the function bends around a point.
  • Evaluate the second derivative at each critical point. If the result is positive, the function is concave up, and there's a relative minimum at that point. If negative, it indicates the function is concave down, signifying a relative maximum.
Consider the same function \(f(x)=3x^{2}-6x+1\). The second derivative is \(f''(x) = 6\), implying that regardless of the value of \(x\), the function remains concave up. Specifically, at \(x=1\), \(f''(1) = 6 > 0\), reinforcing that there's a relative minimum at this point. The second derivative test simplifies checking these critical points by emphasizing concavity.
Critical Points
Critical points are the backbone of discovering a function's local maxima and minima. They're the points where the function's first derivative is either zero or undefined, indicating potential peaks or troughs.
  • To find critical points, calculate the first derivative and set it to zero. Solve for the values of \(x\) where this occurs.
  • Critical points don't automatically mean there's a max or min. It's essential to further analyze these points using either the first or second derivative test to understand their nature.
In the exercise \(f(x)=x^{3}-3x+3\) results in critical points at \(x=\pm1\) when resolving \(f'(x)=3x^{2}-3=0\). Evaluating these further with both the first and second derivative tests will reveal whether these points stand as a relative maxima, minima, or if they're merely flat spots where the slope varies precisely through zero. Understanding critical points is essential when sketching or interpreting graphs to predict how a function behaves.

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Most popular questions from this chapter

In each part: (i) Make a conjecture about the behavior of the graph in the vicinity of its \(x\) -intercepts. (ii) Make a rough sketch of the graph based on your conjecture and the limits of the polynomials as \(x \rightarrow+\infty\) and \(x \rightarrow-\infty .\) (iii) Compare your sketch to the graph generated with a graphing utility. (a) \(y=x(x-1)(x+1) \quad\) (b) \(y=x^{2}(x-1)^{2}(x+1)^{2}\) (d) \(y=x(x-1)^{5}(x+1)^{4}\) (c) \(y=x^{2}(x-1)^{2}(x+1)^{3}\)

Use a graphing utility to make a conjecture about the relative extrema of \(f,\) and then check your conjecture using either the first or second derivative test. $$f(x)=10 \ln x-x$$

In each part sketch a continuous curve \(y=f(x)\) with the stated properties. (a) \(f(2)=4, \quad f^{\prime}(2)=0, \quad f^{\prime \prime}(x)<0\) for all \(x\) (b) \(f(2)=4, f^{\prime}(2)=0, f^{\prime \prime}(x)>0\) for \(x<2, f^{\prime \prime}(x)<0\) for \(x>2\) (c) \(f(2)=4, \quad f^{\prime \prime}(x)>0\) for \(x \neq 2\) and \(\lim _{x \rightarrow 2^{+}} f^{\prime}(x)=-\infty$$\lim _{x \rightarrow 2^{-}} f^{\prime}(x)=+\infty\)

In each part sketch a continuous curve \(y=f(x)\) with the stated properties. (a) \(f(2)=4, f^{\prime}(2)=0, f^{\prime \prime}(x)>0\) for all \(x\) (b) \(f(2)=4, f^{\prime}(2)=0, f^{\prime \prime}(x)<0\) for \(x<2, f^{\prime \prime}(x)>0\) for \(x>2\) (c) \(f(2)=4, f^{\prime \prime}(x)<0\) for \(x \neq 2\) and \(\lim _{x \rightarrow 2^{+}} f^{\prime}(x)=+\infty\) \(\lim _{x \rightarrow 2^{-}} f^{\prime}(x)=-\infty\)

Sketch the graph of the rational function. Show all vertical, horizontal, and oblique asymptotes. $$\frac{(x-2)^{3}}{x^{2}}$$
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