MATH 234, Calculus for BusinessLecture 7, Textbook Sections 3.4Defining the DerivativeFebruary 8th, 2017Announcements• HW 3 is due Thursday, 2/9, 11:59pm; late HWs are not accepted.• Practice exams (and complete solutions) are posted on Compass.• Information for Exam 1 will be posted later this week (stay tuned).ReviewDefinition. The average rate of change of f (x) from x = a to x = b iswhich is also the of the line through .Example 1. Let f(x) =12x2−14x. Find the avg. r.o.c. from x = 1 to x = 1 + h.Solution.Section 3.4, Definition of the DerivativeDefinition (Version 1). The instantaneous rate of change of f(x) at x = a isThis is also called the derivative of f(x) at x = a.Definition (Version 2). The instantaneous rate of change of f(x) at x = a isThis is also called the derivative of f(x) at x = a.1Remark. Both versions will give the same answer.Example 2. Let f(x) =12x2−14x. The avg. r.o.c. from x = 1 to x = 1+h isFind f0(1), which is the instantaneous rate of change at x = 1.Solution Using Version 1.Solution Using Version 2.Example 3 (Review). Let f (x) =12x2−14x. The avg. r.o.c. from x = 1 to x = 1+h isFind the average rate of change from x = 1 to x = 2.Solution.Definition. The line connecting is called the secant line.Slope of the secant line is .In Example 3, the secant line connects .Example 4. Let f (x) =12x2−14x. The avg r.o.c from x = 1 to x = 2 isFind the equation of the secant line from (1, f(1)) to (2, f (2)).Solution.2Example 5. Let f (x) =12x2−14x. The avg. r.o.c. from x = 1 to x = 1+h isFind the equation of the secant line from (1, f(1)) to1 + h, f(1 + h).Solution.For h = 1, As h gets smaller,As h goes to 0, the secant line from (1, f (1)) to (1+h, f(1+h)) becomes the .Definition. The tangent line to the graph of y = f (x) at the point (a, f(a)) is(Write defn. here)The equation of this tangent line isExample 6. Let f (x) =12x2−14x. Recall, .Equation of tangent line at (1, f(1)) isAs h gets smaller, Tangent lineGraphical Interpretation: The tangent lineand does notDo more examples, for mastery.3Example 7. Let f (x) = x2. Compute f0(a) for any value of a.Solution.Example 8. Let f (x) =√x. Compute f0(a) for any positive value of a.Solution.Example 9. Let f (x) =√x. Then f0(a) = .Find the equation of the tangent line to y = f (x) at (4, f (4)).Solution.4Example 10. Let f (x) = x2. From Slide 13, f0(a) = 2a.(1) Find the equation of the tangent line to y = f (x) at (0.5, f(0.5)).(2) Find the equation of the tangent line to y = f (x) at (−1.5, f(−1.5)).Solution.The graph of the function f(x) = x2, along with the tangent lines from (1) and (2) isExample 11 (... from previous slide). Let f (x) = x2. From before, f0(a) =(1) f0(0.5) = , and f(x) = x2is .(2) f0(−1.5) = , and f(x) = x2is .Theorem 1 (As Derivative = Instantaneous Rate of Change...). For a function f (x)(1) If f0(a) > 0, then f (x) is(2) If f0(a) < 0, then f (x) is5WARNING! (The Derivative May Not Always Exist). Let f(x) = |x| =x if x ≥ 0−x if x < 0Does f0(0) = limh→0f(h) − f (0)hexist?If the graph , then .Question 1. What does f0(a) = 0 imply for f (x) at x = a?Remark. This concept will be explored in the future, in
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