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.
(a) Local min at .
(b) Local max at .
(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 .