Solutions to homework 3:

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#2--The appropriate graph is 1.4.33. It is of exponential type, with the limit of 2 as x goes to infinity. Also the y-intercept is (0,1). Note the function can be written f(x) = 2 - 3{^-x} = 2 - 1/3{^x} and since 3{^x} becomes arbitrarily large as x becomes large, the reciprocal of 3{^x} goes to zero as x becomes large We will describe this behavior further when we study limits of functions as x goes to infinity.

#4--The appropriate graph is 1.4.32 since the function is periodic with mzximum value 2 and minimum value 0.

#16--f(x) = x{^1/2} and g(x) = cos(x) gives f(g(x)) = (cos(x)){^1/2}, defined for all x such that cos(x) is non-negative. This includes the interval from -{pi}/2 to {pi}/2 and all intervals obtained by adding integral multiples of 2{pi}. Also g(f(x)) = cos(x{^1/2}), defined for all non-negative x.

#20 f(x) = 1 - x{^2}, g(x) = sin(x), f(g(x)) = 1 - (sin(x)){^2} = (cos(x)){^2}, defined for all real x, and g(f(x)) = sin(1 - x{^2}), also defined for all real x.

#26 If h(x) = (4x - 6){^4/3}, we may choose f(x) = x{^4/3} and g(x) = 4x - 6 ot, slightly less useful, f(x) = x{^1/3} and g(x) = (4x-6){^4}. We will want to use such a setup when we differentiate the function h(x) to get f'(g(x))g'(x) which in this case will be (4/3)(4x-6){^1/3}(4).

#30 If h(x) = 1/(1 + x + x{^2}){^3}, then we may use f(x) = x{^-3} and g(x) = (1 + x + x{^2}). The derivative, by the generalized power rule that shows up in week 4, will be (-3)(1 + x + x{^2}){^-4}(1 + 2x).

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#2 Since (sin(x)){^2}/ x{^2} = (sin(x)/x){^2}, this quantity goes to 1 as x goes to 0.

#4 Since tan(x)/x = (sin(x)/cos(x))(1/x) = (sin(x)/x)(1/cos(x)) this quantity goes to 1*1 = 1 as x goes to zero.

#18 Since (sin(x){^2}/x = (sin(x)/x)sin(x), the first quantity goes to 1 and the second to 0 so the product goes to 0.

#26 Since sin(1/x{^2}) is squeezed between -1 and 1, when we multiply by the non-negative quantity x{^2}, we squeeze x{^2} sin(1/{x}^2} between -x{^2} and x{^2}. Since both of these go to 0 as x goes to 0, so does sin(1/x{^2}).

#62 The function f(x) = x if x is not an integer and f(x) = 0 if x is an integer has a limit at each value a. In particular if a = n, and integer, we have the limit as x goes to n from below = n = limit as x goes to n from above. Note that this common limit, from below and from above, is not equal to the value of the function at any integer except zero. Thus (in the language of the next section) the function is continuous at all points except the non-zero integers.

#66 If f(x) = greatest integer in x/2, then we obtain a step function which makes a jump of 1 at every even integer 2n. These are the only places where the limit as x approaches 2n from below will be 2n-1 while the limit as x approaches 2n from above is 2n, a different value.

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#46 If f(x) = x + 1 when x is less than 1 and 3-x when x is greater than 1, then f is continuous at all points where it is defined. It is not defined at x = 1 so it cannot be continuous there (in the terminology of Edwards and Penney). If we define f(1) = 2, then the function is continuous at x = 1 as well so the discontinuity is removable.

#50 If f(x) = 2x + c for x less than or equal to 3 and 2c - x for x greater than 3, then the limit of f(x) as x goes to 3 from below is 6 + c and the limit as x goes to 3 from above iw 2c - 3. The funtion will be continuous at x = 3 exactly when these two values are equal, i.e. 6 + c = 2c - 3, or c = 9.

#60 The value of the continuous function f(x) = x{^3} -3x{^2} + 1 is negative at -3 and -2 but positive at at -1 so there is a root between -2 and -1 by the Intermediate Value Property. The value at 0 is negative so there is another root in the interval [-1,0]. The valu changes from negative to positive on the interval [2,3] so there is a third root there. A cubic cannot have more than three roots, so there is exactly one root in each of these three intervals and no further root.

Thomas Banchoff
Last modified: Thu Sep 26 17:09:53 EDT 2002