Derivatives - Application of IVT

Introduction

• The Intermediate Value Theorem (IVT) states that if a function is continuous on a closed interval ([a, b]) and takes on values (f(a)) and (f(b)), then it also takes on every value between (f(a)) and (f(b)) at least once.
• In this lesson, we will explore how the IVT can be applied to find solutions for various problems involving derivatives.

Problem 1: Rolle’s Theorem

• Rolle’s Theorem is a special case of the Mean Value Theorem (MVT) that applies when a function (f(x)) has certain properties:
  1. (f(x)) is continuous on the closed interval ([a, b]).
  2. (f(x)) is differentiable on the open interval ((a, b)).
  3. (f(a) = f(b)).
• The theorem states that there exists at least one number (c) in the interval ((a, b)) such that (f’(c) = 0).

Problem 1: Example

• Let’s consider the function (f(x) = x^3 - 3x^2 + 2x) on the interval ([-1, 3]).
• To apply Rolle’s Theorem, we need to check the three properties.
  1. (f(x)) is continuous on the closed interval ([-1, 3]) as it is a polynomial function.
  2. (f(x)) is differentiable on the open interval ((-1, 3)) as it is a polynomial function.
  3. (f(-1) = 0) and (f(3) = 0).
• Therefore, we can conclude that there exists at least one number (c) in the interval ((-1, 3)) such that (f’(c) = 0).

Problem 2: Increasing and Decreasing Functions

• The derivative of a function can help us determine the intervals where the function is increasing or decreasing.
• If the derivative (f’(x)) is positive on an interval ((a, b)), then the function (f(x)) is increasing on that interval.
• If the derivative (f’(x)) is negative on an interval ((a, b)), then the function (f(x)) is decreasing on that interval.
• The points where the derivative changes sign (from positive to negative or from negative to positive) are called critical points.

Problem 2: Example

• Consider the function (f(x) = x^2 - 4x + 3).
• To determine where (f(x)) is increasing or decreasing, we need to find the derivative (f’(x)) and identify its sign.
• Differentiating (f(x)), we get (f’(x) = 2x - 4).
• Setting (f’(x) > 0) gives us (2x - 4 > 0), which simplifies to (x > 2).
• Therefore, (f(x)) is increasing on the interval ((2, \infty)).
• Setting (f’(x) < 0) gives us (2x - 4 < 0), which simplifies to (x < 2).
• Therefore, (f(x)) is decreasing on the interval ((-\infty, 2)).

Problem 3: Optimization

• The derivative of a function can help us find the maximum or minimum values of the function.
• To find the maximum or minimum values of a differentiable function (f(x)) on a closed interval ([a, b]), we need to consider the following:
  1. Find all critical points of (f(x)) in the interval ((a, b)) by setting (f’(x) = 0) or (f’(x)) does not exist.
  2. Find the values of (f(x)) at these critical points and the endpoints (a) and (b).
  3. The maximum value of (f(x)) is the largest value among these values, and the minimum value of (f(x)) is the smallest value.

Problem 3: Example

• Let’s consider the function (f(x) = x^3 - 3x^2 + 2x) on the interval ([-1, 3]).
• To find the maximum and minimum values of (f(x)), we need to follow the steps mentioned earlier.
• Step 1: Finding the critical points by setting (f’(x) = 0).
  • (f’(x) = 3x^2 - 6x + 2)
  • Setting (f’(x) = 0) gives us (3x^2 - 6x + 2 = 0)

Problem 3: Example (contd.)

• Solving the quadratic equation (3x^2 - 6x + 2 = 0), we get (x = \frac{1}{3}) and (x = 2).
• Step 2: Finding the values of (f(x)) at the critical points and endpoints.
  • (f(-1) = -2)
  • (f(\frac{1}{3}) = \frac{2}{27})
  • (f(2) = 0)
  • (f(3) = 8)
• Step 3: Comparing the values to find the maximum and minimum values.
  • Maximum value: 8
  • Minimum value: -2

Conclusion

• By applying the Intermediate Value Theorem, Rolle’s Theorem, and the concepts of increasing/decreasing functions and optimization, we can solve various problems related to derivatives.
• These techniques are particularly useful for determining critical points, finding maximum/minimum values, and understanding the behavior of functions.

Derivatives - Application of IVT

• Problem 4: Fermat’s Theorem

  • Fermat’s Theorem states that if a function (f(x)) has a local maximum or minimum at a point (c), and if (f’(c)) exists, then (f’(c) = 0).
  • This theorem helps us identify critical points where the derivative is zero.

• Problem 4: Example

  • Consider the function (f(x) = x^4 - 4x^2) on the interval ([-2, 2]).
  • We can find the critical points by setting (f’(x) = 0).
  • The derivative (f’(x) = 4x^3 - 8x).
  • Setting (f’(x) = 0) gives us (x(x^2 - 2) = 0).
  • Solving for (x), we get (x = 0) and (x = \pm \sqrt{2}).

• Problem 5: Concavity

  • The second derivative of a function can help us determine its concavity.
  • If (f’’(x) > 0), the function is concave up.
  • If (f’’(x) < 0), the function is concave down.
  • Points where the concavity changes (from concave up to concave down or vice versa) are called inflection points.

• Problem 5: Example

  • Consider the function (f(x) = x^3 - 6x^2 + 9x).
  • To determine the concavity, we need to find the second derivative (f’’(x)).
  • Differentiating (f’(x)), we get (f’’(x) = 6x - 12).
  • Setting (f’’(x) > 0) gives us (x > 2), indicating that the function is concave up on the interval ((2, \infty)).
  • Setting (f’’(x) < 0) gives us (x < 2), indicating that the function is concave down on the interval ((-\infty, 2)).

• Problem 6: Tangent Line Equation

  • The equation of the tangent line to a curve at a given point can be found using the derivative.
  • If the point is ((a, f(a))) and the derivative at that point is (f’(a)), then the equation of the tangent line is given by (y - f(a) = f’(a)(x - a)).

• Problem 6: Example

  • Consider the function (f(x) = x^2 + 2x - 1).
  • To find the equation of the tangent line at the point ((1, 2)), we need to find (f’(1)).
  • Differentiating (f(x)), we get (f’(x) = 2x + 2).
  • Evaluating (f’(1)), we get (f’(1) = 4).
  • Using the point-slope form, the equation of the tangent line is (y - 2 = 4(x - 1)).

• Problem 7: Curve Sketching

  • The first and second derivatives of a function can help us sketch its graph.
  • Points where the first derivative is zero or undefined are possible locations for local extrema or inflection points.
  • The concavity of the function can be determined by examining the sign of the second derivative.
  • The general shape of the graph can be determined by considering the behavior of the function for large values of x.

• Problem 7: Example

  • Consider the function (f(x) = x^3 - 12x).
  • To sketch the graph, we need to find the critical points, inflection points, and analyze the concavity.
  • The derivative (f’(x) = 3x^2 - 12).
  • Setting (f’(x) = 0) gives us (x^2 - 4 = 0).
  • Solving for (x), we get (x = -2) and (x = 2).
  • The second derivative (f’’(x) = 6x), which is always positive, indicating that the function is concave up.
  • The graph of (f(x)) can be sketched using this information.

• Problem 8: Related Rates

  • Related rates problems involve finding the rate at which one quantity changes with respect to another.
  • The key to solving related rates problems is to set up an equation involving the rates of change and the variables in the problem.
  • After setting up the equation, we differentiate both sides with respect to time and solve for the unknown rate.

• Problem 8: Example

  • A conical tank is being filled with water at a constant rate. The height of the water is increasing at a rate of 3 cm/min.
  • The cone has a radius of 4 cm and a maximum height of 10 cm.
  • The goal is to find the rate at which the water level is rising when the height is 6 cm.
  • By setting up an equation involving the rates of change and the variables, we can differentiate both sides and solve for the unknown rate.

Problem 9: Newton’s Method

• Newton’s method is an iterative numerical method used to find the roots of a given function.
• It relies on the concept of linear approximation to iteratively improve an initial guess of a root.
• The formula for Newton’s method is given by: (x_{n+1} = x_n - \frac{f(x_n)}{f’(x_n)})
• This process is repeated until a desired level of accuracy is achieved.
• Newton’s method can be used to solve equations that cannot be solved analytically.

Problem 9: Example

• Let’s say we want to find the root of the equation (x^3 - 4x + 1 = 0).
• We can use Newton’s method to iteratively find the solution.
• Step 1: Choose an initial guess for the root, let’s say (x_0 = 2).
• Step 2: Calculate (f(x_n)) and (f’(x_n)) at each iteration.
• Step 3: Use the formula (x_{n+1} = x_n - \frac{f(x_n)}{f’(x_n)}) to update the guess.
• Step 4: Repeat steps 2 and 3 until a desired level of accuracy is achieved.
• By following these steps, we can find the root of the equation using Newton’s method.

Problem 10: L’Hôpital’s Rule

• L’Hôpital’s Rule is used to evaluate limits involving indeterminate forms.
• An indeterminate form occurs when the limit of a function cannot be easily determined.
• The rule states that if the limit of a ratio of two functions (f(x)) and (g(x)) is in an indeterminate form such as (\frac{0}{0}) or (\frac{\infty}{\infty}), then the limit of the ratio is equal to the limit of the derivative of (f(x)) divided by the derivative of (g(x)) as (x) approaches a given value.
• L’Hôpital’s Rule can be applied to evaluate limits involving functions that are difficult to directly analyze.

Problem 10: Example

• Let’s consider the limit (\lim_{x \to 0} \frac{x - \sin(x)}{x^3}).
• Direct substitution of (x = 0) into the expression gives an indeterminate form (\frac{0}{0}).
• We can use L’Hôpital’s Rule to simplify the expression.
• Taking the derivative of the numerator and the denominator, we get (\lim_{x \to 0} \frac{1 - \cos(x)}{3x^2}).
• Substituting (x = 0) into the simplified expression gives (\lim_{x \to 0} \frac{1 - \cos(0)}{3(0)^2} = \frac{1}{0}).
• Therefore, L’Hôpital’s Rule allows us to evaluate this limit and determine that it is undefined.

Problem 11: Mean Value Theorem

• The Mean Value Theorem (MVT) is a fundamental result of calculus.
• It states that if a function (f(x)) is continuous on a closed interval ([a, b]) and differentiable on the open interval ((a, b)), then there exists at least one number (c) in the open interval ((a, b)) such that: (f’(c) = \frac{f(b) - f(a)}{b - a})
• Geometrically, the MVT states that there exists a point on the graph of (f(x)) where the tangent line is parallel to the secant line connecting the endpoints.

Problem 11: Example

• Consider the function (f(x) = x^2) on the interval ([-1, 1]).
• The function is continuous and differentiable on the interval.
• We can apply the Mean Value Theorem to find the value of (c) that satisfies the given condition.
• Calculating the derivatives and the values of (f(x)) at the endpoints, we have:
  • (f’(-1) = -2)
  • (f’(1) = 2)
  • (f(-1) = 1)
  • (f(1) = 1)
• The slope of the secant line is (\frac{f(1) - f(-1)}{1 - (-1)} = \frac{1 - 1}{2} = 0).
• Therefore, there exists at least one number (c) in the interval ((-1, 1)) where the derivative of (f(x)) is equal to 0, as given by the Mean Value Theorem.

Problem 12: Second Derivative Test

• The Second Derivative Test is used to determine the concavity of a function at critical points.
• It states that if the second derivative (f’’(x)) is positive at a critical point (c), then the function has a local minimum at (c).
• If the second derivative (f’’(x)) is negative at a critical point (c), then the function has a local maximum at (c).
• If the second derivative (f’’(x)) is zero at a critical point (c), then the test is inconclusive.
• The Second Derivative Test helps us analyze the behavior of functions and identify the nature of critical points.

Problem 12: Example

• Consider the function (f(x) = x^3 - 3x^2 + 2x) on the interval ([-1, 3]).
• We have already determined the critical points using Rolle’s Theorem: (x = -1) and (x = 2).
• To apply the Second Derivative Test, we need to find the value of the second derivative (f’’(x)) at these critical points.
• Calculating the second derivative (f’’(x)), we get (f’’(x) = 6x - 6).
• Evaluating (f’’(-1)), we get (f’’(-1) = -12), indicating that (f(x)) has a local maximum at (x = -1).
• Evaluating (f’’(2)), we get (f’’(2) = 6), indicating that (f(x)) has a local minimum at (x = 2).