Solutions to Practice Problems Tuesday, October 28, 2008

Transcription

1 Solutions to Practice Problems Tuesay, October 28, The graph of the function f is shown below. Figure 1: The graph of f(x) What is x 1 + f(x)? What is x 1 f(x)? An oes x 1 f(x) exist? If so, what is it, if not, why? Solution: As x approaches 1 from the left, we re looking at part A of the graph above. On this part of the graph, f(x) goes towar the open circle at 2 as x increases towar 1, so we say that x 1 f(x) 2. Similarly, when x approaches 1 from the right, we re looking at part B of the graph above - the horizontal line segment at y 2. On this part of the graph, f(x) remains constantly at 2 as x ecreases towar 1, so we say x 1 + f(x) 2. Since the right an left it are the same, x 1 f(x) oes exist an x 1 f(x) 2. What is x 1 + f(x)? What is x 1 f(x)? An oes x 1 f(x) exist? If so, what is it, if not, why? Solution: As above, when x approaches 1 from the right, we re on part C of the graph, which goes to 2 as x ecreases to 1. Hence, x 1 + f(x) 2. When x approaches 1 from the left, we re on part B of the graph again. Now, as x increases towar 1, f(x) remains constantly at 2. An so again, x 1 f(x) 2. Once again, since the right an left its agree, the it x 1 f(x) exists an is equal to 2. What is x 3 + f(x)? What is x 3 f(x)? An oes x 3 f(x) exist? If so, what is it, if not, why? Solution: This time, as x approaches 3 from the right, we re on part D of the graph. In this part of the graph, f(x) increases to 2 as x ecreases to 3, hence x 3 + f(x) 2. 1

2 Solutions to Practice Problems Tuesay, October 28, 2008 But when x approaches 3 from the left, we re on part C of the graph again. On this part of the graph, f(x) increases towar 6 as x increases towar 3, hence x 3 f(x) 6. This time the right an left its are ifferent an so the it x 3 f(x) cannot exist. At which (if any) of these three points is f continuous? Solution: In orer to be continuous at a, we must have that x a f(x) f(a) (which means x a f(x) ha better exist, an f(x) ha better be efine at a!). In the first part, x 1 f(x) 2, but f( 1) 3, so f is not continuous at x 1. In the secon part x 1 f(x) 2 AND f(1) 2, so f is continuous at x 1. In the last part, x 3 f(x) oesn t even exist, so f can t possibly be continuous at x Define what it means for a function to have a vertical or horizontal asymptote. Describe what this means for its graph an ientify the vertical an horizontal asymptotes on the graph below. Solution: We say f(x) has a vertical asymptote at x a if f(x) goes to either plus or minus infinity as x gets near a (on either sie). Using it notation: f(x) has a vertical asymptote at x a if x a + f(x) ± or x a f(x) ± (if both the right an left it are the same, then we woul have either x a f(x) or x a f(x) ). Graphically, this means that the curve y f(x) gets very close to the line x a as x gets close to a. In the graph above, x 1 serves as a vertical asymptote for the graph. We say f(x) has a horizontal asymptote at y L if f(x) gets close to L as x goes to either plus or minus infinity. Using it notation: f(x) has a horizontal asymptote at y L if x f(x) L or if x f(x) L (or both). Graphically, this means that the curve y f(x) gets very close to the line y L as x goes of to ±. In the graph above, y 1 serves as a horizontal asymptote. 3. Calculate the it x 2 (x 2 2x + 1/x). What property allows us to o this easily? Solution: Let s call the function above f. So, f(x) x 2 2x + 1/x. Now, since f is a rational function with 2 D f (2 is in the omain of f), the property of Direct Substitution 2

3 Solutions to Practice Problems Tuesay, October 28, 2008 allows us to just plug in the value 2. So, x 2 (x2 2x + 1/x) f(2) /2 1 2 x 4. Calculate the it 2 1 x 1. What property allows us to o this easily? x 1 Solution: Again, let s call our function f. So, f(x) x2 1. Then since the numerator, x 1 x 2 1, is x 2 1 (x 1)(x + 1), we have f(x) x2 1 x 1 (x 1)(x + 1) x 1 x + 1, for x 1 Let s call this function g. So, g(x) x + 1. Then f(x) g(x) except at x 1 (where f is not efine), an so by the property on the top of page 103, x 2 1 x 1 x 1 (x + 1) 2. x 1 5. Suppose f(x), g(x) an h(x) are efine as below. An suppose you know that x f(x) x h(x) 3. Without making any calculations an without using the nice rule we 8 learne on October 20, what is x g(x)? What relationship between f, g an h allows us to raw this conclusion an what theorem are we using? What line serves as a horizontal asymptote to g(x)? f(x) 3x2 2x + 1 8x 2 7 g(x) 3x2 2x + 3 8x 2 7 h(x) 3x2 2x + 8 8x 2 7 Solution: Since f(x) g(x) h(x), an x f(x) x h(x) 3, the Squeeze 8 Theorem tells us that x g(x) x f(x) x h(x) 3. Hence, the line y serves as a horizontal asymptote to g(x). 6. Now let f(x) x2 + 6x + 8 x 2 + 8x + 16 g(x) x2 + 5x + 4 x 2 + 8x + 16 h(x) x2 + 7x + 12 x 2 + 8x + 16 Fin a relationship between these three functions that allows you to use the same theorem as in the previous problem. Use this relationship to give a relationship between x f(x), 3

4 Solutions to Practice Problems Tuesay, October 28, 2008 x g(x) an x h(x). Evaluate each of these its. Do these functions have any horizontal asymptotes? Solution: This time we have g(x) f(x) h(x), which means that x g(x) x f(x) x h(x). Since x g(x) x h(x) 1, the Squeeze Theorem says we must also have that x f(x) 1. An yes, these functions each have y 1 as a horizontal asymptote. 7. You shoul know the it laws backwars an forwars. Fill in the blanks: (f(x) + g(x)) f(x) + g(x) x a x a x a f(x) g(x) (f(x) g(x)) x a x a x a x a f(x) x a g(x) x a f(x) x a g(x) (f(x) g(x)) f(x) g(x) x a x a x a c x a f(x) x a c f(x) x a c x a a x a (f(x))r ( f(x)) r x a as long as the enominator is not 0 8. What oes it mean (in terms of the formal efinitions) if f(x) is continuous at x a? What if I replace continuous with right- or left-continuous? Describe what this means for the graph of the function. Solution: We say f(x) is continuous at x a if x a f(x) f(a). We say f(x) is right-continuous at x a if the right it of f(x) as x approaches a is f(a), that is if x a + f(x) f(x). Similarly, f(x) is left-continuous at x a if the left it of f(x) as x approaches a is f(a), that is if x a f(x) f(x). One way to escribe continuity graphically is to say that we can raw the function near an at a without lifting our pens off the paper - this means there are no holes or jumps. Similarly, a function is right-continuous at x a if we on t nee to lift our pen off the paper as we sketch the graph of f(x) moving towar x a from the right (so, we re rawing the graph starting to the right of a an moving towar it) in orer to reach the value f(a). Likewise for the left. For example, the piecewise function in Figure 1 (from Problem 1) is right-continuous at 3 but not left-continuous. 9. You shoul also know how continuity is affecte by those it laws. Fill in the blank: If f an g are continuous at a an if c is some constant, then (1) f + g, (2) f g, (3) c f, (4) f g an (5) f are continuous (as long as g(a) 0). g Choose one of (1)-(5) above an write explicitly what this means (in terms of its). Use the it laws to justify the equality. 4

5 Solutions to Practice Problems Tuesay, October 28, 2008 (1) If f + g is continuous at a, we must have x a (f(x) + g(x)) f(a) + g(a). The it laws tell us that x a (f(x) + g(x)) x a f(x) + x a g(x), an since f an g are assume to be continuous at a, x a f(x) f(a) an x a g(x) g(a) an so we have that x a (f(x) + g(x)) f(a) + g(a), as esire. The others are similar. 10. The graphs of the functions f an g are shown below. (Note: The scale on g s graph is a little o.) Figure 2: The graph of f(x) Figure 3: The graph of g(x) Solution Without trying to figure out the functions involve, we know that (f g)(x) 2 x 1 Since g is continuous at 1 an f is continuous at g(1) 1, because this allows us to apply Theorem 9 on page 125, which tells us that f g is continuous at 1 (hence, x 1 (f g)(x) (f g)(1) f(g(1)) f(1) 2). 5

6 Solutions to Practice Problems Tuesay, October 28, 2008 In general, the theorem states that if g is continuous at a value x a an f is continuous at g(a), then the composite function f g is continuous at x a. 11. Let f(x) 1. Where oes f have a horizontal asymptote? What about g(x) x 2 f(x + 1) 1 an h(x) f(x) x2? What oes this mean, in terms of (x+1) 2 x 2 x 2 its? Solution: We know from Theorem 5 on page 133 (an hopefully from our euctive abilities) that f(x) has a horizontal asymptote at y 0. Amittely, the theorem actually states that the it of the function as x approaches ± is 0, but by our efinition of a horizontal asymptote, these are really the same. Now, the graph of g(x) is just the same as f(x) except that it is shifte one unit to the left. But shifting the graph of f(x) to the left oesn t actually chance the functions behavior at ±, hence y 0 is also a horizontal asymptote for g(x). On the other han, the graph of h(x) is the graph of f(x) shifte up one unit, an this oes affect the behavior of the function at ± since it increases the value of f(x) everywhere. So y 1 is a horizontal asymptote for this function. In terms of its: an h(x) x ± g(x) x ± (f(x) + 1) x ± f(x + 1) 0 x ± f(x) x ± x ± 12. Justify the labele steps using the it laws an theorems covere in class an in your textbook. 3x 2 2x + 7 x 2x 2 x + 3 A x ( ) 3x 2 2x+7 x ( 2 2x ) 2 x+3 x x x 2 x x x 2 B x ( x x ( x ) x 2 ) x 2 C 1 x 3 2 x + 7 x x 1 x 2 1 x 2 x + 3 x x 1 x 2 D Step A can be justifie by the property on the top of page 103, which states that when two functions ar Step B can be justifie by it law number 5 on page 99. Step C can be justifie by it laws 1, 2 an 3 on page 99. Step D can be justifie by Theorem 5 on page

7 Solutions to Practice Problems Tuesay, October 28, We foun a nice rule in class for cases like the one above. We sai that if f(x) a n x n + a n 1 x n a 1 x + a 0 an g(x) b m x m + b m 1 x m b 1 x + b 0 f(x) are polynomials (an we assume a n 0 an b m 0), then the it x is the same g(x) as the it of what other (much simpler) function? f(x) x g(x) a n x n x b m x m This means that we have the following, a n b m f(x) x g(x) a n b m if n > m if n m 0 if n < m a Note: n b m shoul really just be ±, but the way we ecie whether it shoul be + or is by looking at the sign (+/-) of the quotient an b m. For example if a n 2 an b m 3 then their quotient is positive, an so the it woul be +. But if a n 2 an b m 3, then their quotient is negative, an the it woul be. 14. Using the rule in 13, evaluate the its: x 4 7x + 4 x x 4 1 an x x 3 x x 2 2 Solution: x 4 7x + 4 x x 4 1 x 4 x x 1 4 x x 3 x x 2 2 x 3 x x Define the erivative of f(x) at x a. (Hint: There are two its you coul use here, but you shoul use at least one of them!) Use this efinition to fin f (2) when f(x) 3x 3 + x (you nee to know how to take this it, so using the shortcuts we learne on October 27 won t help you prepare for your exam). Describe what this number is in terms of the graph (Hint: you shoul use the wor tangent). Solution: In terms of the graph, the erivative of a function at x a is the slope of the tangent line to y f(x) at (a, f(a)). The first time we saw the efinition of the erivative it came from an exercise which showe us that the slope of the tangent line can be seen as the it of the slopes of secant lines originating from (a, f(a)). This iea gave us the following: 7

8 Solutions to Practice Problems Tuesay, October 28, 2008 x f(a) f f(x) f(a) (a) x a x a Notice that the thing we re taking the it of is actually just the slope of the secant line through the points (a, f(a)) an (x, f(x)), an taking this it has the effect of moving x closer to a, which moves the secant line closer to the tangent line. We also saw an equivalent efinition of the erivative, which we like because it allowe us to later fin a general formula for the erivative of a function. This efinition is: x f(a) f f(a + h) f(a) (a) h 0 h If f(x) 3x 3 + x, then f(2) , so we have: f (2) h 0 3(2 + h) 3 + (2 + h) ( 22) h h 0 3( h h 2 + h 3 ) h + 22 h h h 18h 2 3h 3 + h + 24 h h 0 35h 18h 2 3h 3 h h h 3h 2 35 So the slope of the tangent line to the graph y 3x 3 + x at (2, 22) is Define the erivative of f(x) as a function of x. Use this efinition to fin f (x) for f(x) 3x 3 + x an verify your answer for f (2). Then fin the equation for the tangent line to f(x) at (2, f(2)). Solution: We efine the erivative of f(x) as a function by taking the it: For f(x) 3x 3 + x, we have: x f(x) f f(x + h) f(x) (x) h 0 h f (x) 3(x + h) 3 + (x + h) ( 3x 3 + x) h 0 h 3x 3 9x 2 h 9xh 2 3h 3 + x + h + 3x 3 x h 0 h 9x 2 h 9xh 2 3h 3 + h h 0 h 9x 2 9xh 3h h 0 9x

9 Solutions to Practice Problems Tuesay, October 28, 2008 Using this formula, f (2) , which is what we got before! 17. Using the same f(x) as in 15 an 16, fin the values of x where f (x) 0. What oes this tell us about the graph of f(x)? Sketch the graph of f (x). On what intervals is f(x) increasing an on what intervals is f(x) ecreasing? Solution: Since we fount that f (x) 9x 2 + 1, we can fin the places where the erivative is 0 by setting this equal to 0 an solving for x: 9x we coul use the quaratic formula, but we on t nee it 1 9x x2 1 9 x ± 1 3 x So, the points ( 1, 2) an ( 1, 2 ) have horizontal tangent lines What oes it mean for a function g to be ifferentiable at a point x a? How can a function fail to be ifferentiable? Give an example of a function which is not ifferentiable at x 1. (A graph will suffice for this last part.) Solution: We efine the erivative of g at x a as a it (see problem 15). We say g is ifferentiable at x a if this it exists. A function can fail to be ifferentiable by being iscontinuous or by being pointy. Both functions in problem 10 are continuous at x 1, but neither is ifferentiable since both are pointy. 19. If the function p(t) 1.5t 2 gives my position (istance in meters away from a fixe point - like a stop light) at time t (measure in secons), fin my velocity v(t) an my acceleration a(t). How fast am I going after 10 secons? At t 10, am I speeing up or slowing own? How can you tell? Solution: Velocity is the erivative of positing/isplacement, so v(t) p (t) 2 1.5t 3t (m/s). Acceleration is the erivative of velocity, i.e. the secon erivative of position/isplacement, so a(t) v (t) p (t) 3 (m/s 2 ). My velocity after 10 secons is given by v(10), so I must be going 30m/s at time t 10 sec. The question of whether I am speeing up or slowing own is really the question: is my velocity increasing or ecreasing. When we want to know whether a function is increasing or ecreasing, we look at it s erivative - in this case, that s my acceleration - if the erivative of a function is strictly positive, then the function is increasing; if it s strictly negative, then it s ecreasing. Since my acceleration is constant, a(10) 2, which is greater than 0, I am still speeing up. 20. Compute the following erivatives (use the shortcuts!). 9

10 Solutions to Practice Problems Tuesay, October 28, 2008 (a) f(x) 4x 3 7x + 2 x 2. So, f (x) 12x 2 7 2x. (b) g(x) 2 e x 1 x 2 2 e x x 2. So, g (x) 2 e x ( 2x 3 ) 2 e x + 2 x 3. (c) h(x) 3 x + 3 x. So, h (x) ln(3) 3 x 3 x The Prouct Rule tells us that: This is the same as: (f(x) g(x)) x 22. The Quotient Rule tells us that: This is the same as: x ( ) x f(x) g(x) + f(x) (f(x) g(x)) f (x) g(x) + f(x) g (x) ( ) f(x) g(x) x g(x) ( x f(x)) g(x) f(x) x g(x) (g(x)) 2 ( ) f(x) f (x) g(x) f(x) g (x) g(x) (g(x)) Use the Prouct an Quotient Rules to compute the following erivatives: (a) ( e x (4x 3 7x + 2 x 2 ) ) x x (4x3 7x + 2 x 2 ) x ex (4x 3 7x + 2 x 2 ) + e x e x (4x 3 7x + 2 x 2 ) + e x (12x 2 7 2x) e x (4x 3 x 2 7x x 2 2x 7) e x (4x x 2 9x 7) (b) ( 2 e x 1 ) x 2 x x + 1 x ( ) 2 e x x 2 x + 1 (2 x ex x 2 ) (x + 1) (2 e x x 2 ) (x + 1) x (x + 1) 2 (2 ex ( 2)x 3 ) (x + 1) (2 e x x 2 ) 1 (x + 1) 2 (2 ex + 2 x 3 ) (x + 1) (2 e x x 2 ) 1 (x + 1) 2 2x ex + 2x x e x + 2 x 3 2 e x + x 2 (x + 1) 2 2x ex + 3 x x 3 (x + 1) 2 10

11 Solutions to Practice Problems Tuesay, October 28, 2008 (c) x ( 3 x + 3 ) x 3 x x ( ) 3 x + 3 x 1 3 x x (3x + 3 x 1 ) (3 x ) (3 x + 3 x 1 ) x 3x (3 x ) 2 (ln(3) 3x + 3 ( 1) x 2 ) (3 x ) (3 x + 3 x 1 ) (ln(3) 3 x ) 3 2x (ln(3) 32x 3 x 2 3 x ) (ln(3) 3 2x + 3 x 1 ln(3) 3 x ) 3 2x ln(3) 32x 3 x 2 3 x ln(3) 3 2x 3 x 1 ln(3) 3 x 3 2x 3 x 2 3 x 3 x 1 ln(3) 3 x 3 2x ( 3 3x )(x 2 + ln(3) x 1 ) 3 2x ( 3)(x 2 + ln(3) x 1 ) 3 x 1 ln(3) x x 2 3 x 1 x () (( 3 x + 3 x ) ) (2x 1) (( 3 x + 3 x 1) (2x 1) ) x ( 3 x + 3 x 1) (2x 1) + ( 3 x + 3 x 1) (2x 1) x x ( ln(3) 3 x + 3 ( 1) x 2) (2x 1) + ( 3 x + 3 x 1) 2 ( ln(3) 3 x 3 x 2) (2x 1) + ( 2 3 x + 6 x 1) ( ln(3) 2x 3 x 6x x 2) ( ln(3) 3 x 3 x 2) + ( 2 3 x + 6 x 1) ln(3) 2x 3 x 6 x 1 ln(3) 3 x + 3 x x + 6 x 1 ln(3) 2x 3 x ln(3) 3 x + 3 x x 24. Given that rules to fin: x sin(x) cos(x) an cos(x) sin(x), use the prouct an/or quotient x 11

12 Solutions to Practice Problems Tuesay, October 28, 2008 (a) x tan(x) x ( ) sin(x) cos(x) sin(x) cos(x) sin(x) cos(x) x x cos(x) cos(x) sin(x) ( sin(x)) (cos(x))2 + 1 (sec(x)) 2 (b) x cot(x) x ( ) cos(x) sin(x) cos(x) sin(x) cos(x) sin(x) x x sin(x) sin(x) cos(x) cos(x) ((sin(x))2 + ) 1 (csc(x)) 2 (c) x sec(x) ( ) 1 x cos(x) (1) cos(x) (1) x x cos(x) 0 cos(x) ( sin(x)) sin(x) sec(x) tan(x) 12

13 Solutions to Practice Problems Tuesay, October 28, 2008 () x csc(x) ( ) 1 x sin(x) (1) sin(x) (1) sin(x) x x 0 sin(x) cos(x) cos(x) csc(x) cot(x) 13