Math 1A – Final Exam Solutions – Fall ’04

 

1.      The graph below shows y = f(x).  a.       State, with reasons, the numbers a at which  does not exist. SOLN:  The limit does not exist at a = –3 since the limit from the left is 1 while the limit from the right is -1.  Also, the limit does not exist at a = 2 since the value of f is unbounded above both from the left and the right. b.      State where the function is discontinuous and classify the discontinuity as removable, jump or asymptote. SOLN:  There is a jump discontinuity at a = –3, a removable discontinuity at a = 1 and a vertical asymptote discontinuity at a = 2. c.       State, with reasons, the numbers at which the function is not differentiable. SOLN:  The function is not differentiable where a = –3, 1 and 2, since these are discontinuities.  The function is not differentiable at a = –2 since the slope is 0 as approached from the left and 2 as approached from the right. The function is no differentiable at a = –1 since the slope is a constant 2 as approached from the left while the slope approaches some negative value from the right.  The function is not differentiable at a = 0 since the tangent line is vertical there.

2.       Suppose f(x) is a function such that for all x > 1, .
What is ?  Why?
SOLN:  Since  approaches 1 from below and  approaches 1 from above, we can conclude that a function caught between these must also converge to 1.

3.      Sketch a graph for a function that meets the given conditions:

4.      The graph of  is shown below. a.       Sketch a graph for .   SOLN:

b.      Sketch a possible graph for .  SOLN (note that only the essential ingredients of increasing/decreasing and concave up/down are important to show here):

5.      Find a linear approximation for the function  at a = 0 and use it to approximate .
SOLN:  In a neighborhood of x = 0,  so that

6.      Find the slope of the line tangent to the curve  at .
SOLN:  Assuming y is a function of x and equating derivatives,
Plugging in the given values:

7.      Find an equation for the line tangent to  at .
SOLN:  Here  and the slope is  so the equation could be

8.      At what point on the curve  is the tangent line horizontal?
SOLN: 

9.      Find the local and global extreme values of the function   on the interval [-1,2].

SOLN:   or x = 0.  Since y’<0 only on ,

y has a global min at the left endpoint  and a local min where x = 2.  y has a global max where .  There is an inflection point where x = 0, but since y is increasing both before and after this point, there is no max nor min at the origin.

10.  Find the point on the hyperbola xy = 16 that is closest to the point (4,0).

SOLN:  A point on the hyperbola can be described by .   The square of the distance from (4,0) to that point is .  Differentiating,

Sketching both the square of the distance and its rate of change per change in x and using the calculator to find the zero of the latter, we see the distance is minimized where x = 5.52111, whence y = 16/5.52111 = 2.898.

11.  Evaluate the limit: .
SOLN:  Since both the numerator and the denominator are differentiable functions approaching zero,

12.  The velocity of a wave of length L in deep water is  where k and C are known positive constants.  What is the length of the wave that gives the minimum velocity?
SOLN: 

13.  What is the maximum slope of a line connecting the origin (0,0) with a point on the parabola ?
SOLN:  .  These correspond to the local max at  and a local min at ,  but the slope is unbounded above, as seen by looking at .

14.  Show that the y-coordinate of the point (x,y) on the curve described by  that is closest to the point (0,2) can be found by solving .  Use Newton’s method to solve the equation.  Hint: Minimizing the distance D is equivalent to minimizing the square of the distance, D².  Also, the curve has origin symmetry, so you can just work with the distance in the first quadrant.
SOLN:  .  If (x,y) is a point on this curve then the square of the distance from (x,y) to (0,2) is .  Differentiating, , and setting the derivative to zero,  and setting the numerator to zero yields the desired equation.  On the TI Voyage 200, you can write a program for Newton’s method (shown at left below).  Calling “Newton(1)” yields the output in the screenshot at right.
           

15.  A metal storage tank with volume V is to be constructed in the shape of a right circular cylinder with a hemispherical top.  What radius and height will require the least amount of metal?  SOLN:  The volume is fixed at . Solving for h: .  Now the surface area is .  Substituting for h in terms of r:

Setting the derivative to zero: