Homework 4 Solutions, 2/2/7

Transcription

1 Homework 4 Solutions, 2/2/7 Question Given that the number a is such that the following limit L exists, determine a and L: x 3 a L x 3 x 2 7x + 2. We notice that the denominator x 2 7x + 2 factorizes as (x 3)(x 4), so goes to zero as x goes to 3. Then the limit cannot exist unless the numerator x 3 a also vanishes as x goes to 3, which entails that 3 3 a = 0, so a = 27. When a = 27, we have the factorization: Then we get: x 3 a = x 3 27 = (x 3)(x 2 + 3x + 9). L x 3 x 3 27 x 2 7x + 2 x 3 (x 3)(x 2 + 3x + 9) (x 3)(x 4) x 3 x 2 + 3x + 9 x 4 = (3) + 9 = = 27. So the required limit exists if and only if a = 27 and then we have L = 27.

2 Question 2 Evaluate the following limits. If a limit does not exist say why. If the limit goes to or to, say why. lim x 4 x 2 6 2x + 3x 3 lim x 4 x 2 6 2x + 3x 3 (x 2 6)( 2x + + 3x 3) x 4 ( 2x + 3x 3)( 2x + + 3x 3) (x 2 6)( 2x + + 3x 3) x 4 ( 2x + ) 2 ( 3x 3) 2 (x 2 6)( 2x + + 3x 3) x 4 2x + (3x 3) (x 2 6)( 2x + + 3x 3) x 4 4 x (x 4)(x + 4)( 2x + + 3x 3) x 4 4 x ( (x + 4)( 2x + + ) 3x 3) x 4 = (4 + 4)( 2(4) + + 3(4) 3) = 8( ) = 8( 9 + 9) = 8(3 + 3) = 8(6) = 48. 2

3 x x lim 2 + x x 2 + 0x + 25 lim x x x 2 + x 2 + 0x + 25 x x 2 ( + x 2 ) x x 2 ( + 0x + 25x 2 ) x x 2 + x 2 x x 2 ( + 0x + 25x 2 ) x( x) + x 2 x x 2 ( + 0x + 25x 2 ) x 2 + x 2 x x 2 ( + 0x + 25x 2 ) + x 2 x + 0x + 25x = =. Here we used that x 2 = x = x, when x is negative and that negative integral powers of x go to zero as x goes to minus infinity. lim x + x 2 + 5x 6 x 2 2x + x 2 + 5x 6 lim x + x 2 2x + x + (x + 6)(x ) (x ) 2 x + x + 6 x =. Here the denominator goes to 0 + and the numerator to 6, so their quotient goes to plus infinity. 3

4 Question 3 Find the horizontal and vertical asymptotes of the function f(x) = 2x2 x x 2 x 2 and use the information of these asymptotes to give a rough plot of the graph of the function. 2x 2 x lim f(x) x x x 2 x 2 x 2 (2 x x 2 ) x x 2 ( x 2x 2 ) 2 x x 2 x ( x 2x 2 ) = = 2, lim f(x) 2x 2 x x x x 2 x 2 x 2 (2 x x 2 ) x x 2 ( x 2x 2 ) 2 x x 2 x ( x 2x 2 ) = = 2. So as x ±, the graph y = f(x) approaches the horizontal asymptote y = 2. In fact we have for the difference between y and 2: y 2 = 2x2 x x 2 x 2 2 = 2x2 x 2(x 2 x 2) x 2 x 2 = 2x2 x 2x 2 + 2x + 4 = x + 3 x 2 x 2 x 2 x 2. Then for large positive x, the numerator x+3 is positive and the denominator x 2 x 2 is also positive, so y is larger than 2, so for x the graph approaches y = 2 from above. However, for large negative x, the numerator x + 3 is negative, whereas the denominator x 2 x 2 is still positive, so y is smaller than 2, so for x the graph approaches y = 2 from below. 4

5 Next we note that x 2 x 2 = (x 2)(x+), so there are vertical asymptotes at x = 2 and x =. We examine the behavior of f(x) as we approach these asymptotes from each side. 2x 2 x lim f(x) x 2 + x 2 + (x 2)(x + ) 2(2 2 ) 2 x 2 + (x 2)(2 + ) 5 x 2 + 3(x 2) =, 2x 2 x lim f(x) x 2 x 2 (x 2)(x + ) 2(2 2 ) 2 x 2 (x 2)(2 + ) 5 x 2 3(x 2) =, lim f(x) 2x 2 x x + x + (x 2)(x + ) 2(( ) 2 ) ( ) x + ( 2)(x + ) lim f(x) 2x 2 x x x (x 2)(x + ) 2(( ) 2 ) ( ) x ( 2)(x + ) Plotting shows in the x-interval (, ), that for large negative x, the graph of y = f(x) starts out just below the horizontal asymptote y = 2, then decreases, and is concave down at first, switching to concave up at ( , ). Then it reaches a local minimum at ( , ), before increasing, crossing the level y = 2 at x = 3 and then increasing, staying always concave up, going to as x approaches from the left. In the x-interval (, 2) the curve is concave down, increasing from as x + crossing the x-axis at ( 2, 0), crossing the y-axis at (0, 2 ), reaching a local maximum at ( , ), then decreasing, crossing the x-axis again at (, 0) and then going down to as x 2. Finally, in the x-interval (2, ), y starts out very large, and decreases, always concave up, approaching the asymptote y = 2 from above, as x. In this region, it never goes below the asymptotic level y = 2. x + 2 3(x 2) =, x 2 3(x 2) =. 5

6 Question 4 Find the equations of the tangent and normal lines to the curve y = x + at the point (3, 2) and sketch the curve and the lines on the same graph. If f(x) = x +, then the slope f (3) of the tangent line at x = 3 is given by the limit: f f(x) f(3) (3) x 3 x 3 x + 2 x 3 x 3 ( x + 2)( x + + 2) x 3 (x 3)( x + + 2) ( x + ) x 3 (x 3)( x + + 2) x + 4 x 3 (x 3)( x + + 2) x 3 x 3 (x 3)( x + + 2) = x 3 x = = 4. So the required tangent line has slope and passes through the point (3, 2), 4 so has the equation: y 2 = (x 3), 4y 8 = x 3, x 4y + 5 = 0. 4 Then the required normal line has slope ( = 4 and passes through the 4) point (3, 2), so has the equation: y 2 = 4(x 3) = 4x + 2, y + 4x = 4. The plot of y = x + is the top half of a parabola (lying above the interval [, ), with axis the x-axis and vertex the point (, 0). The parabola is increasing and concave down for all x. The tangent line lies above the parabola, touching it only at (3, 2). The tangent line has intercepts (0,.25) and ( 5, 0). The normal line has intercepts (0, 4) and (3.5, 0). 6

7 Question 5 The function y = f(x) is defined by the following formulas: If x < 0, then f(x) = sin(x) x If 0 x 2, f(x) = 3 x 2 x + If 2 < x, f(x) = 2 x. Sketch the graph of f. Discuss (with explanation) the continuity of f. Estimate (with explanation) the range of the function f. It is clear that the function is continuous everywhere except possibly at x = 0, or at x = 2. The graph of sin(x) for x negative is an oscillation of decreasing amplitude where the amplitude goes to zero as x. x Plotting the oscillation, we see that its lowest point occurs at the first minimum, which we estimate to be the point ( , ). Also as x 0, the graph (also to be proved in class) indicates that lim x 0 f(x) sin(x) x 0 =. x Next we have: lim x 0 + f(x) x 0 +(3 x 2 x +) = = + = = f(0). So lim x 0 f(x) = = f(0) and the function f is continuous at x = 0. Next we plot f over the interval [0, 2]: the graph rises steadily and curves concave up from (0, ) to (2, ) = (2, 6). lim x 2 f(x) x 2 (3 x 2 x +) = = 9 4+ = 6 = f(2). Also lim x x = 2 2 = 6 = f(2). So the function f is continuous at x = 2 also. The last part of the graph is the portion of an hyperbola, descending from the point (2, 6) and curving concave up, approaching the horizontal asymptote y = 0 from above, as x. In particular the absolute maximum of f is 6 at x = 2 and the absolute minimum is about at the point x = So the function f is continuous everywhere, with range [ , 6]. 7