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Calculus Quiz: Limit Definition of a Derivative Instructions: This quiz consists of multiple-choice questions and short-answer questions related to the limit definition of a derivative in calculus. Attempt all questions. Good luck! Multiple Choice Questions: 1. What does the limit definition of a derivative describe? a) The rate of change of a function at a given point b) The area under a curve c) The integral of a function d) The slope of a tangent line to a curve at a given point 2. Which of the following statements represents the limit definition of the derivative of a function f(x) at a point a? a) f′(a)=lim x→af(x)x−af′(a)=limx→ax−af(x) b) f′(a)=lim h→0f(a+h)−f(a)hf′(a)=limh→0hf(a+h)−f(a) c) f′(a)=lim h→0f(h)−f(a)h−af′(a)=limh→0h−af(h)−f(a) d) f′(a)=lim x→af(a)−f(x)xf′(a)=limx→axf(a)−f(x) 3. What is the geometric interpretation of the limit definition of the derivative? a) The area under the curve b) The slope of the secant line between two points on the curve c) The slope of the tangent line to the curve at a given point d) The length of the curve Short Answer Questions: 4. Define the limit definition of the derivative of a function f(x) at a point a. 5. Use the limit definition of the derivative to find the derivative of the function f(x)=3x2+2x−1f(x)=3x2+2x−1 at the point x = 2. 6. Explain why the limit definition of the derivative involves taking the limit as the interval around the point of interest approaches zero. Answers: 1. d) The slope of a tangent line to a curve at a given point 2. b) f′(a)=lim h→0f(a+h)−f(a)hf′(a)=limh→0hf(a+h)−f(a) 3. c) The slope of the tangent line to the curve at a given point Short Answer Solutions: 4. The limit definition of the derivative of a function f(x) at a point a is given by: f′(a)=lim h→0f(a+h)−f(a)hf′(a)=limh→0hf(a+h)−f(a) 5. Using the limit definition, f′(x)=lim h→0f(x+h)−f(x)hf′(x)=limh→0hf(x+h)−f(x) Substituting f(x)=3x2+2x−1f(x)=3x2+2x−1, we get:
f′(2)=lim h→03(2+h)2+2(2+h)−1−(3(2)2+2(2)−1)hf′(2)=limh→0h3(2+h)2+2(2+h)−1−(3(2)2 +2(2)−1) f′(2)=lim h→03(4+4h+h2)+4+2h−1−(12+4−1)hf′(2)=limh→0h3(4+4h+h2)+4+2h−1−(12+4− 1) f′(2)=lim h→012+12h+3h2+4+2h−1−15hf′(2)=limh→0h12+12h+3h2+4+2h−1−15 f′(2)=lim h→015h+3h2hf′(2)=limh→0h15h+3h2 f′(2)=lim h→0(15+3h)f′(2)=limh→0(15+3h) f′(2)=15f′(2)=15 6. The limit definition of the derivative involves taking the limit as the interval around the point of interest approaches zero because it represents the slope of the tangent line to the curve at that point. As the interval around the point shrinks to zero, the secant line becomes closer to the tangent line, providing an accurate approximation of the slope of the tangent.
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