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Chapter 4: Problem 4

Explain how to apply the Second Derivative Test.

Short Answer

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Question: Explain the steps to apply the Second Derivative Test to classify a critical point as a local maximum, local minimum, or a saddle point. Answer: To apply the Second Derivative Test, follow these steps: 1. Find the first derivative, f'(x), of the given function using differentiation rules. 2. Determine the critical points by setting f'(x) equal to 0 and solving for x. 3. Calculate the second derivative, f''(x), by differentiating f'(x) using differentiation rules. 4. Plug each critical point into the second derivative, f''(x), and evaluate. 5. Use the Second Derivative Test to classify the critical points: If f''(x) > 0, the point is a local minimum; if f''(x) < 0, it is a local maximum; if f''(x) = 0 or doesn't exist, the test is inconclusive and alternative methods should be used.

Step by step solution

01

Find the first derivative

To apply the Second Derivative Test, first, we need to find the first derivative of the given function, denoted by f'(x). Use rules for differentiation, such as the power rule, product rule, and chain rule, to determine the derivative.
02

Find the critical points

Critical points occur when the first derivative is equal to zero or doesn't exist. Set f'(x) equal to 0, and then solve for x to find the critical points. These are the values of x where the function could have a local maximum, local minimum, or a saddle point.
03

Find the second derivative

Next, find the second derivative, denoted by f''(x), by differentiating the first derivative, f'(x). Just as with the first derivative, use rules for differentiation, such as the power rule, product rule, and chain rule, to calculate the second derivative.
04

Plug critical points into the second derivative

Substitute each critical point found in Step 2 into the second derivative, f''(x), and evaluate.
05

Use the Second Derivative Test to classify the critical points

Now, we will use the Second Derivative Test. If f''(x) > 0 at a critical point, the function is concave up at this point, and the critical point is a local minimum. If f''(x) < 0, the function is concave down, and the critical point is a local maximum. If f''(x) = 0 or the second derivative doesn't exist at the critical point, the Second Derivative Test is inconclusive, and we cannot determine whether the critical point is a local maximum, local minimum, or a saddle point using this test alone. In such cases, we can use the First Derivative Test or other methods to analyze the critical point.

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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 key points on a function where the behavior of the graph changes. These points occur where the first derivative, denoted as \( f'(x) \), equals zero or does not exist. This is because, at critical points, the slope of the tangent to the curve is zero, indicating a possible maximum, minimum, or saddle point.

To find critical points:
  • Determine the first derivative \( f'(x) \) of the function.
  • Solve the equation \( f'(x) = 0 \).
  • Identify points where \( f'(x) \) is undefined.
Once you have these points, they become the candidates for local extrema or points of inflection, which require further analysis using the second derivative or other methods.
First Derivative
The first derivative of a function, \( f'(x) \), represents the rate of change or the slope of the function at any given point. It tells us how fast or slow a function is increasing or decreasing. Calculating \( f'(x) \) is a vital step when analyzing a function's behavior as it helps determine where critical points occur.

To find the first derivative, apply differentiation rules:
  • Power Rule: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
  • Product Rule: If \( f(x) = u(x)v(x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \).
  • Chain Rule: If \( y = g(f(x)) \), then \( y' = g'(f(x)) \cdot f'(x) \).
Each of these rules helps break down more complex functions into manageable parts for differentiation.
Second Derivative
The second derivative, \( f''(x) \), is the derivative of the first derivative \( f'(x) \). It provides insight into the concavity of a function and thus helps classify critical points more accurately. The second derivative test uses \( f''(x) \) to determine if a critical point is a local minimum, local maximum, or if the test is inconclusive.

To find \( f''(x) \):
  • Differentiate \( f'(x) \) using applicable differentiation rules.
  • Plug critical points into \( f''(x) \) to analyze changes in concavity.
If \( f''(x) > 0 \) at a critical point, the function is concave up and has a local minimum there. If \( f''(x) < 0 \), the function is concave down, indicating a local maximum. If \( f''(x) = 0 \), further investigation is required.
Differentiation Rules
Differentiation rules are fundamental tools in calculus that allow us to find derivatives of various functions. Understanding these rules makes it easier to calculate both the first and second derivatives, essential for applying the Second Derivative Test.

Some primary differentiation rules include:
  • Constant Rule: The derivative of a constant is zero.
  • Sum Rule: The derivative of a sum equals the sum of the derivatives.
  • Difference Rule: The derivative of a difference equals the difference of the derivatives.
  • Quotient Rule: Used when dividing two functions, \( f(x) = \frac{u(x)}{v(x)} \), with \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \).
By mastering these rules, differentiation becomes a straightforward process, allowing you to analyze and interpret the behavior of functions accurately.

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Most popular questions from this chapter

Graph several functions that satisfy the following differential equations. Then find and graph the particular function that satisfies the given initial condition. $$f^{\prime}(s)=4 \sec s \tan s ; f(\pi / 4)=1$$

Show that any exponential function \(b^{x}\), for \(b>1,\) grows faster than \(x^{p},\) for \(p>0\)

Show that \(f(x)=\log _{a} x\) and \(g(x)=\) \(\log _{b} x,\) where \(a>1\) and \(b>1,\) grow at a comparable rate as \(x \rightarrow \infty\)

An eigenvalue problem A certain kind of differential equation (see Section 7.9 ) leads to the root-finding problem tan \(\pi \lambda=\lambda\) where the roots \(\lambda\) are called eigenvalues. Find the first three positive eigenvalues of this problem.

Find the function \(F\) that satisfies the following differential equations and initial conditions. $$F^{\prime \prime}(x)=\cos x, F^{\prime}(0)=3, F(\pi)=4$$
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