MATH 151 Engineering Mathematics I

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1 MATH 151 Engineering Mathematics I Spring 2019, WEEK 10 JoungDong Kim Week 10 Section 4.2, 4.3, 4.4 Mean Value Theorem, How Derivatives Affect the Shape of a Graph, Indeterminate Forms and L Hospital s Rule. Section 4.2 The Mean Value Theorem The Mean Value Theorem (MVT) Let f be a function that satisfies the following hypotheses: 1. f is continuous on the closed interval [a,b]. 2. f is differentiable on the open interval (a, b). Then there is a number c in (a,b) such that f (c) = f(b) f(a) b a or, equivalently, f(b) f(a) = f (c)(b a) Graphically, this means the tangent line to the graph of f(x) at x = c is parallel to the secent line joining the points (a,f(a)) and (b,f(b)). 1

2 Ex.1) Given f(x) = 4 x 2, show f(x) satisfies the mean value theorem on the interval [1,2] and find all c that satisfy the conclusion of the mean value theorem. Ex.2) Suppose 1 f (x) 4 for all x in interval [2,5]. Prove 3 f(5) f(2) 12. 2

3 Section 4.3 How Derivatives Affect the Shape of a Graph Increasing/Decreasing Test (a) If f > 0 on an interval I, then f is increasing on I. (b) If f < 0 on an interval I, then f is decreasing on I. The First Derivative Test Suppose that c is a critical number of a continuous function f. (a) If f changes from positive to negative at c, then f has a local maximum at c. (b) If f changes from negative to positive at c, then f has a local minimum at c. 3

4 Ex.3) Find all intervals of increase and decrease and identify all local extrema. a) f(x) = x 4 +4x 3 b) f(x) = x x+1 c) f(x) = xe 2x 4

5 Definition. If the graph of f lies above all of its tangents on an interval I, then it is called concave upward on I. If the graph of f lies below all of its tangents on an interval I, then it is called concave downward on I. OR, If the slopes of a curve become progressively larger as x increases, then we say f is concave upward. If the slopes of a curve become progressively smaller as x increases, then we say f is concave downward. Concavity Test and Inflection points (a) If f > 0 for all x in I, then f is increasing, hence f is concave upward on I. (b) If f < 0 for all x in I, then f is decreasing, hence f is concave downward on I. (c) If f changes concavity at x = c, and x = c is in the domain of f, then x = c is a inflection point of f. 5

6 Ex.4) Use the given graph of f to find the following. (a) The intervals on which f is increasing. (b) The intervals on which f is decreasing. (c) The intervals on which f is concave upward. (d) The intervals on which f is concave downward. (e) The coordinates of the points of inflection. 6

7 Ex.5) The graph of the second derivative f of a function f is shown below. (a) On what intervals is f concave upward? (b) On what intervals is f concave downward? (c) Stat the x-coordinates of the points of inflection. 7

8 Ex.6) The graph of the derivative f of a function f is shown. (a) On what intervals is f increasing or decreasing? (b) At what values of x does f have a local maximum or minimum? (c) On what interval is f concave upward or downward? (d) State the x-coordinates of the points of inflection. (e) Sketch a graph of f. 8

9 Ex.7) If f (x) = e x2, what can be said about f? 9

10 The Second Derivative Test If x = c is a critical number for f(x), then (a) If f (c) > 0, then f is concave up, therefore f(x) has a local minimum at x = c. (b) If f (c) < 0, then f is concave down, therefore f(x) has a local maximum at x = c. (c) If f (c) = 0 or does not exist, then the test fails, therefore use the first derivative test to find the local extreme. Ex.8) Use the second derivative test to find the local extrema for f(x) = x 3 3x 1. 10

11 Ex.9) Sketch a graph of f satisfying the following conditions. (a) f (x) > 0 on the interval (,1) and f (x) < 0 on the interval (1, ). (b) f (x) > 0 on the interval (, 2) and (2, ). (c) f (x) < 0 on the interval ( 2,2). (d) lim f(x) = 2 and lim f(x) = 0. x x 11

12 Ex.10) Sketch the graph of f(x) = x 4 4x 3 by locating intervals of increase/decrease, local extrema, concavity, and inflection points. 12

13 Ex.11) Sketch the graph of f(x) = x ln x by locating intervals of increase/decrease, local extrema, concavity, and inflection points. 13

14 x Ex.12) Sketch the graph of f(x) = by locating intervals of increase/decrease, local extrema, (x 1) 2 concavity, and inflection points. For your convenience, f (x) = x 1 (x 1) 3, f (x) = 2x+4 (x 1) 4. 14

15 Section 4.4 Indeterminate Forms and L Hospital s Rule Indeterminate form If we have a limit of the form f(x) lim x a g(x) = 0 0 or, then this limit may or may not exist and is called an indeterminate form. L Hospital s Rule Suppose f and g are differentiable and g (x) 0 on an open interval I that contains a (except possibly at a). Suppose that limf(x) = 0 and lim g(x) = 0 x a x a or that limf(x) = ± and limg(x) = ±, x a x a (In other words, we have an indeterminate form of type 0 0 or ) then f(x) lim x a g(x) = lim f (x) x a g (x) if the limit on the right side exists (or is or ). f(x) NOTE If lim x a g(x) = 0 or 0 Rule., the limit is NOT indeterminate. CANNOT use L Hospital s 15

16 Ex.13) lim x 1 lnx x 1 Ex.14) lim x 0 2 x 1 x Ex.15) lim x 0 sinx x x 3 Ex.16) lim x 0 sinmx sinnx (lnx) 3 Ex.17) lim x x 2 lnx Ex.18) lim x 2 2 x 16

17 Indeterminate Products If lim x a f(x)g(x) = 0, this limit is an indeterminate product. We can deal with it by writing the product fg as a quotient: fg = f 1/g or fg = g 1/f. This converts the given limit into an indeterminate form of type 0 0 or L Hosptal s Rule. so that we can use Ex.19) lim x 0 +xlnx Ex.20) lim x 0 + xsecx Ex.21) lim x 1 +(x 1)tan ( π 2 x ) 17

18 Indeterminate Difference If lim x a (f(x) g(x)) =, this limit is an indeterminate difference. Ex.22) lim (secx tanx) x π 2 ( 1 Ex.23) lim x 1 lnx 1 ) x 1 ( 2x+1 Ex.24) lim x 0 + sinx 1 ) x 18

19 Indeterminate Power If lim x a f(x) g(x) is of the form 0 0, 0, or 1, then the limit is an indeterminate power. To solve such a limit, take the natural logarithm. Ex.25) lim x x 3/x Ex.26) lim x 0 +(1+sin4x)cotx 19

20 ( ) 2x+1 2x+3 Ex.27) lim x 2x+5 20

Calculus 1 Math 151 Week 10 Rob Rahm. Theorem 1.1. Rolle s Theorem. Let f be a function that satisfies the following three hypotheses:

Calculus 1 Math 151 Week 10 Rob Rahm 1 Mean Value Theorem Theorem 1.1. Rolle s Theorem. Let f be a function that satisfies the following three hypotheses: (1) f is continuous on [a, b]. (2) f is differentiable