Math 1300: Calculus I, Instructor: Dana Ernst 

Partial Solutions to Review #2

Here are some partial solutions to the questions on Review for Midterm #2.  For most of the problems, I’ve just provided the answer, so that you can check to make sure that you are doing the problem correctly.  In some cases, I’ve provided some of the intermediate steps to help guide you.

 

1. The function
   
    is a good example.

 

2. The function
   
    is a good example.  Here,
   
   .

 

3. The function
   
    is a good example.  Here,
   
   .

 

4. Use appropriate derivative formulas to get the following.

 

(a)    

(b)   

(c)    

(d)   

(e)    

(f)    

(g)    

(h)   

(i)     

 

5. Use
   
   ,
   
   , and
   
    to get
   
   .

 

6. Use
   
   ,
   
   , and
   
    to get
   
   .

 

7. First, find the derivative using the product rule:
   
   .  The x-values that make this function 0 determine the horizontal tangents.  These values are
   
   .  The x-values where this function is undefined determine the possible vertical tangents.  These values are
   
   .  In fact, these do determine vertical tangents.

 

8. Use implicit differentiation to get the following.

(a)    

(b)   

 

9. First, find the derivative:
   
   .  The only critical point of this function in the interval
   
    is
   
   .  Testing this point and the endpoints of the interval, we see that the max occurs at
   
    and the minimum occurs at
   
   .

 

10.  First, find the derivative:
    
    .  The critical points of this function are
    
    .  But neither of them is in the interval
    
    . So we only need to test the endpoints of the interval.  We see that the max occurs at
    
     and the minimum occurs at
    
    .

 

11. The answer is
    
    .  Hint: Minimize the distance squared instead of the distance; this eliminates the square root in the distance formula.

 

12. See Book.

 

13. See Book.

 

14. In each case, draw a number line and put 2 down as the only critical number.  Then figure out whether the function is increasing or decreasing on either side of
    
    .

(a)   Local min at .

(b)  Local max at .

(c)   Neither.

 

15. In each case, check to see if the 3 hypotheses of Rolle’s Theorem are satisfied.

(a)   No, g is not continuous at , which is in the interval .

(b)  Yes, h is continuous on  (the only discontinuities are at , which are not in the interval),  is differentiable on  (to see this, find  and observe that the only discontinuities of this function are at , as well), and .  Note: According to the book, we should have , but this is unnecessary.

 

16.  f satisfies the hypotheses of the MVT since (i) f is continuous on  (the only discontinuity of f is at , which is not in the interval), and (ii) f is differentiable on .  To find all points that satisfy the conclusion of the MVT, solve for x in

.

            You will find out that the answer is .  Note that  is not in the interval.

 

17.  If we implicitly differentiate with respect to t, we get .  Plug in  (decreasing!) and , and then solve for  meters/sec.

 

18.  Here is the equation you need to use: .  At some moment in time, we have ,  (since distance between observer and point on the ground below the nugget is decreasing).  Differentiating with respect to t, we get: .  We can see that we are also going to need an x-value.  To get this, plug  into  and then solve for x.  You should get .  Now, plug in all of the information that you have and solve: .

 

19.  To determine the intervals of increase and decrease, do the usual number line thing with the first derivative.  To determine the intervals of concave up and down, do the number line thing with the second derivative.

(a)   For , notice that the domain is .  f is decreasing on , f is increasing on , and f is concave up on .

(b)  g is decreasing on  and , g is increasing on , g is concave up on  and , and g is concave down on .

 

20.  Same game as #19.

(a)   Local min at , inflection points at .

(b)  Along the way, you should get  and .  By setting  equal to 0 and solving for x, you should get that the only x-value in the interval is .  Doing the usual number line thing, you determine that g has a local min at .  Next set  equal to 0 and solve for x by factoring.  You will encounter  and , which become  and , respectively.  Solving for x in each, you will get that the only x-values in the interval are  and .  Using a calculator, you can pick test points on the number line.  You will find out that g has inflection points at .

 

21.  Along the way you will get the following: x-intercept: , y-intercept: , vertical asymptote: None, horizontal asymptote: None, g is increasing on , g is decreasing on , , , g is concave down on  and , g is concave up on , and .  Use this information to sketch graph.

 

22.  Find the second derivative: .  Then plug in 2: .

 

23.  Here, you are just supposed to find the equation of the tangent line to  at .  The answer is .

 

24.  First, get the linear approximation for  using .  You get .  Then .

 

25.  To find , just find the derivative, then tack on .

(a)    

(b)