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Sketch the graph of a twice-differentiable function y=f(x) that passes through the points (-2,2), (-1,1), (0,0),(1,1), and (2,2) and whose first two derivatives have the following sign patterns.
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Ch. 4 - Applications of Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 4, Problema 71
Let ƒ(x) = 3x - x³ . Show that the equation ƒ(𝓍) = -4 has a solution in the interval [2,3] and use Newton’s method to find it.
Guida verificata passo dopo passo1
First, verify that the function ƒ(x) = 3x - x³ has a solution in the interval [2,3] by applying the Intermediate Value Theorem. Calculate ƒ(2) and ƒ(3) to check if there is a sign change.
Calculate ƒ(2): ƒ(2) = 3(2) - (2)³ = 6 - 8 = -2. Calculate ƒ(3): ƒ(3) = 3(3) - (3)³ = 9 - 27 = -18. Since ƒ(2) = -2 and ƒ(3) = -18, there is no sign change, so check the values of ƒ(x) at the endpoints to ensure a solution exists.
Since the function is continuous and ƒ(2) = -2 and ƒ(3) = -18, check if the value -4 lies between these two values. Since -4 is between -2 and -18, by the Intermediate Value Theorem, there is at least one solution in the interval [2,3].
To apply Newton's method, start with an initial guess x₀ within the interval [2,3]. A reasonable choice is x₀ = 2.5. Newton's method formula is xₙ₊₁ = xₙ - ƒ(xₙ)/ƒ'(xₙ).
Find the derivative ƒ'(x) = 3 - 3x². Use this derivative in the Newton's method formula: xₙ₊₁ = xₙ - (3xₙ - xₙ³ + 4)/(3 - 3xₙ²). Substitute x₀ = 2.5 into this formula to find the next approximation x₁.
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Intermediate Value Theorem
The Intermediate Value Theorem states that if a continuous function takes on two values at two points, it must take on every value between those two points at least once. In this case, we can evaluate ƒ(2) and ƒ(3) to show that the function changes sign over the interval [2,3], indicating that there is at least one solution to the equation ƒ(𝓍) = -4 within that interval.
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Fundamental Theorem of Calculus Part 1
Newton's Method
Newton's Method is an iterative numerical technique used to find approximate solutions to equations. It starts with an initial guess and refines it using the formula x_{n+1} = x_n - ƒ(x_n)/ƒ'(x_n), where ƒ' is the derivative of ƒ. This method is particularly effective for finding roots of functions when the derivative is known and can lead to rapid convergence to the actual solution.
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Derivatives
The derivative of a function measures how the function's output changes as its input changes. It is a fundamental concept in calculus that provides information about the function's slope and behavior. For Newton's Method, calculating the derivative of ƒ(x) is essential, as it is used to determine the slope at the current approximation, guiding the next guess towards the root.
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