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is false a critical number does not have to be where a local maximum or local minimum occurs. Fermat's Theorem suggests that we should start looking for extreme values of f at the numbers c where f'(c) = 0 and f'(c) does not exist 4.1: Maximum and Minimum Values 1 Find critical numbers for the following s(t) = t4 + 4t³ +21² s' (t) = 4t³ + 12t² + 4t = 0 4t (t² + 3t+1) = 0 t = 0, ܂ Find the critical numbers for the following Find critical numbers for the following 4.1: Maximum and Minimum Values −3 ± √5 2 f(z)=xre* f'(x) = xe + et = 0 et (x + 1) = 0 Find critical numbers for the following x = -1 g(0) = 0 + sin g'(0) = 1 + cos 0 = 0 cos=-1 θ = π + 2πη, nel g(t) = √t(1 t) g(t) = t - t 3 g' (t) = t - t = 0 2 t(1-3t] = = 1/3,0 t = = 0 The extreme values of a continuous function f on a closed interval [a, b] always exists; they occur either at a, at b, or at a critical number of f in (a,b). Use the Closed Interval Method to find absolute maximum and minimum values Steps 2 1. Find values of f at critical numbers in (a, b) 2. Find values of f at endpoints 3. Choose the largest and smallest values from the results in steps 1 and 2 Find the absolute maximum and minimum values of f on [0,3] f(x)= x³ 3x + 1 f'(x)= 3x²-3 = 0 3x² = 3 f(0) = 1 Find the absolute maximum and minimum values of f on [1,2] f(1) = f(0) = 0 x = 1 f(1) = -1 Since there is no critical numbers on this graph between 1 and 2, the absolute maximum is 2 and the absolute minimum is 1 1 4.1: Maximum and Minimum Values f(x) = Find the absolute maximum and minimum values on f on [0, 1] f(x)=e*e-2x x x + 1 f'(x) =-e-² +2e-2x = 0 -p-2 [1-2e] = In e = In In 1/2 = 0 -z=1²/ ln 2 -x = ln 1- In 2 x = ln 2 f(3) = 19 f(2)= 23 3 4.1: Maximum and Minimum Values f(ln 2) 1/14 f(1) = 1-1/2 e² 4 4.2: Mean Value Theorem Rolle's Theorem: If a function f 1. is continuous on [a, b] 2. is differentiable on (a, b) 3. has f(a) f(b) Then there is at least one number c = (a, b) such that f'(c) = 0. That is, there is at least one point where the slope of the tangent line is zero between the endpoints Example: Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle's Theorem f(x)= x³ 3x² + 2x + 5, [0,2] Since f is a polynomial it is continuous on [0,2] and differentiable on (0,2) f(0) = 5 f(2)= 5 So, f(0) = f(2). So by Rolle's Theorem 3c (0,2) so that f'(c) = 0 f'(c) = 3c²6c+ 2 = 0 6± √36-4(3) (2) 6 4.2: Mean Value Theorem C = 3+√3 3 3± √√3 3 → 0.4226, 1.5774 Mean Value Theorem: is similar to Rolle's Theorem except we do not assume that f(a) equals f(b). 1 If the function f: 1. is continuous on [a,b] 2. is differentiable on (a, b) Then there is at least one number c (a, b) such that Notice (a, b) is an open interval f'(c) = f(b)-f(a) b-a Since is the slope of the line joining the endpoints of the graph, the Mean Value Theorem says there is at least one point on the graph somewhere between the endpoints where the tangent line has the same slope as the line joining the endpoints Example: Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval and then find all the numbers c that satisfy the conclusion of the theorem 4.2: Mean Value Theorem f(b) - f(a) b-a f(x) = x³ 3x² + x, [0,3] .. by the MVT 3cE (0,3) so that f'(c) f(3)-f(0) 3-0 = 3c²6c+1= 3-0 = 1 3 0 3c²6c0 3c(c - 2) c = 0,2 Example: Let f(x) = (2+1). Show that there is no value of c in (0,2) such that f(2) – f(0) = f'(c)(2-0). Why does this not contradict the Mean Value Theorem? 2 = ƒ(2) f(0) = f'(c)(2 - 0) f(2)-f(0) = f'(c) 2-0 3+1 (c-1)-(c+1) 2 (c − 1)² 2 (c-1)² -2=2(c-1)² -1 = (c − 1)² 2 no c value exists. position function s = f(t), then the average distance traveled and the velocity at t = c is b-a time elapsed If an object moves in a straight line with velocity between t = a and t = b is f(b)-f(a) f'(c) = 0. Thus, the MVT tells us that at some time t = c between a and b, the instantaneous velocity, f'(c), is equal to the average velocity. For instance, if a car traveled 180 km in 2 hours, then the speedometer must have read 90 km/hr at least once Example: David is driving on an interstate highway which has a speed limit of 55 mph. At 2 PM he is at milepost 110 and at 5 PM he is at milepost 290. Is this enough evidence to prove that David is guilty of speeding? 180 miles 3 hrs Yes, this is enough evidence to prove that David was speeding Example: Suppose that f'(x) ≤2 for all x. If f(1) = 8 what is the largest possible value that f(5) could be? f(5) - 8 5-1 f(5) 88 f(5) ≤ 16 4.2: Mean Value Theorem 60 mi/hr Example: Show that the equation 3x - 2 + cos (2) = 0 has exactly one real root. f is the sum of a cos and a polynomial, so f is continuous everywhere (-7) f(2)= 6 - 2 + COS π = 3 Since -50<3, IVT applies and a root on (-1,2) Assume a 2nd root exists Since f is also differentiable by the MVT, if a second root exists, then f'(c) = 0 ≤2 f(-1)=-3-2 + cos f'(c)=3-s sin 6 π ㅠ 3 = sin (c) 2 = sin (c) -c) = 0 ¹ (c) = -5 Which is unsolvable and would contradict MVT .. a 2nd root does not exist 3 4.3: How Derivatives Affect the Shape of the Graph Increasing Decreasing Test: a. If f'(x) > 0 for all x on an interval, then f is increasing on that interval b. If f'(x) <0 on an interval, then f is decreasing on that interval Example 1: Where is f(x) = x² + 6x increasing? f'(x) = 2x + 6 (-3,00) Example 2: Where is f(x) = x³ - 3x² increasing and where is it decreasing? f'(x) = 3x² - 6x Increasing: (-∞,0) U (2,00) Decreasing: (0,2) 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, f has a local maximum at c. b. If f' changes from negative to positive at c, f has a local minimum at c. c. If f' does not change sign at c (that is, f' is positive pn both sides of c, or negative on both sides of c), then f has no local maximum or minimum at c. Example 3: Find where the local maximum and minimum values of f(x) = 1-2 sin x occur f'(x) = -2 cos x = 0 4.3: How Derivatives Affect the Shape of the Graph 1 Maximum: x = 3 + 2πn, n = I Minimum: x=2+2mn, n = I x = 2 + πη, η ει Concavity Test: a. If f'(x) > 0 for all a in I, then the graph of f is concave up on I b. If f(x) < 0 for all x in I, then the graph of f is concave down on I. Definition: A point P on a curve is called an inflection point if the curve changes from concave upward to concave downward or from concave downward to concave upward at P 4.3: How Derivatives Affect the Shape of the Graph The Second Derivative Test: Suppose f" is continuous near c a. If f'(c) = 0 and f"(c) > 0, then f has a local minimum at c b. If f'(c) = 0 and f"(c) <0, then f has a local maximum at c 2 4.4: Indeterminate Forms and L'Hospital's Rule L'Hospital's Rule: Suppose f and g are differentiable and g'(x) 70 near a (except possibly at a) Suppose that limx→a f(x) = 0 and limx→a g(x) = 0 or that limx→a f(x) = ±∞o and limx→a g(x) = ∞ f(x) g(x) Then the limz+a f(x) or limx→a f = is called an indeterminate form Thus by L'Hospital's Rule L'Hospital Applies Evaluate the following f(x) lim x→a g(x) if the right side exists, or is too Evaluate the following (this is not proper form) x²7x+12 4.4: Indeterminate Forms and L'Hospital's Rule f'(x) x→a g'(x) = lim lim x 3 lim x 3 x² - 9 2x - 7 2x = x + tan x lim x 0 sin a 1 6 1 Using L'Hospital's, Evaluate the following So, using L'Hospital's Evaluate the following So, using L'Hospital's Evaluate the following Evaluate the following lim x-0 lim x + tan x = 0 x-0 x + tan x sin a 4.4: Indeterminate Forms and L'Hospital's Rule lim sin = 0 x-0 lim x-0 et ∞+I lim xx x lim lim I→∞ I lim e = ∞o 1 + sec² x lim x = ∞ 81X lim €48 COS a lim e = ∞ In (In x) X lim ln(lnx) = ∞ x →∞ 1 lim xxx ln x lim x = ∞ →∞ x + tan 2x tan 2x 0 cos 3x lim *cos 7x = 1 lim sec 7x cos 3x x→ 2 1 = 2 2 So, using L'Hospital's Evaluate the following So, using L'Hospital's Using L'Hospital's again Evaluate the following let y = x lim cos 3x = 0 4.4: Indeterminate Forms and L'Hospital's Rule lim cos 7x = 0 x→ -3 sin (3x) 3 lim x-7 sin (7x) = 7 lim x-1 (-1) In x-1- ln x lim x-1 (In x)(x - 1) lim - 1- ln x = 0 x→1 x1¹x)(x-1)=0 x-1- ln x 1 1 lim = lim xi (In x)(x - 1) →1 ln x + -1 lim 1 x = 0 x+1 lim ln x + 1-¹=0 x→1 x-2 lim 2-1 1 +x lim x +0+* In y = x ln x In y lim = +0+* 1 In x x lim ln x = -∞ x→0+ 2-1 3 Using L'Hospital's Rule Evaluate the following Let y = (e + x)² Using L'Hospital's Rule 1 lim = ∞ T-01 x 4.4: Indeterminate Forms and L'Hospital's Rule In y = lim z 0+ In y lim x 2-0 y = 1 lim (e* + x) = xx lim In y lim ln(e+ x) x x+x ∞0+x e²+x lim 1 (ex + 1) 1 e +1 x + x² ∞0+2 So, if you keep taking the natural log of e to the power of x, you keep getting infinity over infinity, so simplify In y = lim 1 ∞0+I y = e 4 4.5: Summary of Curve Sketching Curve Sketching Checklist: 1. Domain of f 2. x-intercepts y-intercepts 3. Symmetry Wrt y-axis Wrt origin Periodic 4. Horizontal Asymptotes Vertical Asymptotes 5. Increasing Intervals Decreasing Intervals. 6. Critical Numbers Local Extrema 7. Concave Upward Concave Downward Inflection Points. 4.5: Summary of Curve Sketching For what is f(x) defined? What are the solutions to f(x) = 0? What is the value of f(0)? Is f(-x) = f(x)? Is f(-x) = -f(x)? Is there a number p such that f(x+p) = f(x) for all in the domain? Does limx→∞ f (x) or lime-o f(x) exist? what intervals is f'(x) > 0? what intervals is f'(x) <0? Where does f'(x) = 0 or not exist? Where are the local maxima and minima? On what intervals is f"(x) ≥ 0? On what intervals is f"(x) <0? Where does f"(x) = 0 or not exist? 1 4.7: Optimization Example 1: Find the dimensions of the largest rectangular field enclosed by 100 feet of fencing A = xy 4.7: Optimization A(x) = x(50-x) y = 50 x 12 - Tr² h = 2πr A'(x) = 2x + 50 = 0 50 2 I = = 25 y = 50-25 y = 25 100 2x + 2y Example 2: A cylindrical aluminum cup is to be made from 12 square inches of aluminum. What is the largest possible volume of such a cup? 12 = 2πrh+ #r² V = πr²h A(x) = x² + 50x h = 6 πT P 2 1 V(r) = πr² 6 πr A' = 4.7: Optimization 3π V'(r) = 6 - x² + y² = 0=6- 10² 4 ㅠ 1 = (100 - x²) (-2x) + (100 – 2²) ² 2 We can't assume this is the max radius, but if we plug in numbers before and after, we see that it changes direction at this point, therefore r = √ is the radius that yields the maximum volume V v (√²) r = 3π 2 =p² Example 3: Find the dimensions of the largest rectangular peg that can be put into a round hole with a diameter 10 cm π A' = .p² = 4.514 in ³ A=x√ 100 - x² -1 (100x²) [x² + 100 - x²] V(r) = 6r — 77³ A' = -x² (100 100 - 2x² √100x² x = 5√2 68.1655y√/2 y 9.640 = 0 A = xy - x²)+(100 - x²) ¹ 2 Example 4: A fence 8 feet tall runs parallel to a tall building at a distance of 4 feet from the building. What is the length of the shortest ladder that will reach the ground over the fence to the wall of the building h 4+y When y6.3496, x has a min 4.7: Optimization 8 Y X = x² =h²+(4+ y)² x = √h² + (4 + y)² 32 + 8y Y h = 2 = √(32 + 8y) ². Y +(4+ y)² 16.648 ft. 3 4.10 Antiderivatives F(x) is an antiderivative of the function, f(x), if F'(x) = f(x). If F is an antiderivative of f, then all other antiderivatives of f have the form F(x)+c, where c is a constant. f(x) Function x" except n=-1 sin a cos a sec² x sec a tanz 1+z² b. f(x) Find all antiderivatives of the following a. f(x)=x7 = cos x - sec² x c. f(x)=x+x-² F(x) Form of all antiderivatives 4.10 Antiderivatives +1 +C n+1 - cos x + C sin x + C tan +C seca + C sin¹x + C tan¹a + C = 28+C F(x) = F(x) = sin x tan x + C F(x) = 2x² - -x-¹+C 1 d. f(x)=x²-√√x³ e. f(0) = e + sec 0 tan 0 F(0)=e+sec 0 + C The general solution to the differential equation dy = f(x) is all antiderivatives of f . To find a particular solution, first find the general solution, then substitute the given values to determine the constant C 3 2 F(x) = ²x³ − ²x³ + C 5 Find the antiderivative of F(x) of f(x) that satisfies the given condition f(x) = 4-3 (1+x²) ¹, where F(1) = 0 Find f(t) if f""(t) = t-√t 4.10 Antiderivatives F(x) = 4x3 tan-¹(x) + C 0 = 4(1) - 3 tan-¹(1) + C 3π 4 0=4- -4+ Find f(x) if f'(x)=√₁2 where f(1) = 1 3π 4 +C 3π F(x) = 4x3 tan¯¹(x) - 4+ 4 = C 1 = 4 sin-1 f(x) = 4 sin¹(x) + C (²). 1= 4( ) + C 1-²=C 3 + C f(x) = 4 sin¹(x) + 1 - 2п 3 f(t)=²=²+c +C 4 f' (t) = {/t³ - 1 t -t² + Cx + D 15 2 1 f(t) = 24t¹ 4.10 Antiderivatives 8 105 C 5² +22² + -x² +Dx+E Find f(x) if f(x) = sinx, f(0) = 1, f'(0) = 1, f"(0) = 1 f"(x) = cos(x) + C 1 = cos(0) = C→C = 2 f"(x) = cos x + 2 f'(x) = sin x + 2x + C 1 = sin(0) + 2(0) + C → C = 1 f'(x)=sin x + 2x + 1 f(x) = cos x + x² + x + c 3