MATH 019: Final Review December 3, 2017

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1 Name: MATH 019: Final Review December 3, Given f(x) = x 5, use the first or second derivative test to complete the following (a) Calculate f (x). If using the second derivative test, calculate f (x) also. (b) Find all critical numbers (c) Find all possible inflection points (d) Write the table for the test here. Indicate which test you are using. (e) On what interval(s) is f(x) concave up? (f) On what interval(s) is f(x) concave down? (g) On what interval(s) is f(x) increasing? (h) On what interval(s) is f(x) decreasing? (i) Report the relative minima in point form (j) Report the relative maxima in point form (k) Report the global minimum in point form (l) Report the global maximum in point form (m) Sketch f(x). Make sure that the extrema and inflection points are positioned accurately. 2. Complete the following sentences with the word function or derivative based on the plot which contains a function and its derivative. (a) The orange curve is the (b) The purple curve is the Q-1

2 3. What is the quadratic formula? 4. Calculate the roots of f(x) = 5x x. 5. Write the following expressions in their simplest form (a) e ln 3 10+e π (b) log 100 (c) log 3 3 φ where φ is any function. This is the Greek letter phi. 6. Compute log π e (use the change of base formula) 7. What is the definition of the derivative (the limit)? 8. Calculate the derivative of f(x) = x algebraically. 9. Given lim f(x) = 4 and lim g(x) = 9 then find (a) lim 3f(x) + 2g(x) f(x) (b) lim 2 ln g(x) 10. What is the definition of a limit? It starts like: Let f be a function and L, a R. If f(x) = L then lim 11. Calculate lim x 3 f(x) where f(x) = 12. What is the definition of continuity for x = c 13. Where is f(x), from question 11, continuous? 14. What is the definition of differentiable? 15. Where is f(x), from question 11, differentiable? 16. Calculate dy dx for the following (a) y = log a x 3 where a R is a constant (b) y = x log 7 x (c) y = 10 e x ln x (d) e y + y = xy (e) y = x 9 2 6x 5 2 x (f) y = 0.8 x { x + 1 x < 3 4 x 3 Q-2

3 17. (Lizards) In the summer, the activity level of a certain type of lizard varies according to the time of day. A biologist has determined that the activity level is given by the function a(t) = 0.008t t t + 7 where t is the number of hours after 12 noon. When is the activity level highest? When is is lowest? 18. A company wishes to run a utility cable from point A on the shore to an installation point B on the island. The island is 6 miles from the shore. Point A is 9 miles from Point C, the point on the shore is closest to Point B. To run the cable underwater, it costs 5/4 the price of running the cable on land (per mile). Assume that the cable starts at A and runs along the shoreline, then angles and runs underwater to the island. Find the point at which the line should begin to angle in order to yield the minimum total cost. (a) Label the diagram. (b) State restrictions on each variable. (c) Construct a function for cost and call it C. (d) Calculate the derivative of C with respect to distance. (e) Find all critical numbers. (f) Write the table for the test of your choice here. (g) Interpret your table. Along what path should the company install the cable? Include units! Q-3

4 Answers 1. Given f(x) = x 5, USE THE SECOND DERIVATIVE TEST! PART (g) AND (h) REQUIRE IT! (a) f (x) = 5x 4 and f (x) = 20x 3 (b) Critical Number: x = 0 (c) Possible Inflection Point: (0, 0) (d) Second Derivative Test (and First Derivative Test since the second is inconclusive for determining if an extremum occurs at x = 0) x f (x) f (x) (e) f(x) is concave up on (0, ) (f) f(x) is concave down on (, 0) (g) f(x) is increasing on (, ) (h) f(x) is never decreasing (i) There are no relative minima (j) There are no relative maxima (k) There is no global minimum (l) There is no global maximum (m) A-1

5 2. Complete the following sentences with the word function or derivative based on the plot which contains a function and its derivative. (a) The orange curve is the function (b) The purple curve is the derivative 3. x = b± b 2 4ac 2a are solutions to 0 = ax 2 + bx + c 4. x = ± , Write the following expressions in their simplest form (a) (b) 2 (c) φ e π 6. log π e = log e log π f f(x+h) f(x) (x) = lim h 0 h 8. f (x) = 2x 9. Given lim f(x) = 4 and lim g(x) = 9 then find (a) 30 (b) 16 ln What is the definition of a limit? It starts like: Let f be a function and L, a R. If f(x) = L then (both of the following are equivalent definitions) lim (1) f(x) can be as close to L as desired for a choice of x sufficiently close to a. (2) for all ε > 0 there exists a δ > 0 such that x a < δ implies that f(x) f(a) < ε. 11. lim x 3 f(x) = 4 A-2

6 12. A function f is continuous at x = c if the following three conditions are satisfied: (1) f(c) is defined (2) lim f(x) exists x c (3) lim f(x) = f(c) x c 13. f(x) is continuous for all real numbers. 14. A function f is differentiable when x = c if the following three conditions are satisfied (1) f is continuous when x = c (2) f is smooth when x = c (3) f does not have a vertical tangent line at x = c 15. f(x) is differentiable everywhere except where x = 3 (not smooth) 16. Calculate dy dx for the following (a) Use the chain rule to get (b) Use the product rule to get dy dx = 3x2 (ln a)x 3 = 3 (ln a)x. ( ) dy 1 dx = x + log (ln 7)x 7 x. (c) Use the quotient rule to get [ ] dy (ln x)e x dx = 10 e x 1 x (ln x) 2 (d) Use implicit differentiation and the product rule to get e y dy dx + dy dx = x dy dx + y dy dx = (e) Simplify the expression and then use the power rule to get [ ] (ln x) = 10e x 1 x (ln x) 2. y e y + 1 x. y = x 9 2 6x 5 2 x = x 4 6x 2 dy dx = 4x3 12x when x 0. Alternatively, use the quotient rule without simplifying the expression (Your answer will not look the same if you do this). (f) Use the exponential rule to get dy dx = 0.8x ln 0.8. A-3

7 17. (Lizards) In the summer, the activity level of a certain type of lizard varies according to the time of day. A biologist has determined that the activity level is given by the function a(t) = 0.008t t t + 7 where t is the number of hours after 12 noon. When is the activity level highest? When is is lowest? This is an optimization problem where t [0, 24] since there are 24 hours in a day. Note that t = 0 and t = 24 are the same time of day. By the extreme value theorem we know that there exists a minimum and a maximum activity level of the lizard. First, we find critical numbers by computing a (t) = 0.024t t By setting this equal to 0, we find the critical numbers are t 5.072, The maximum and minimum will occur at either the endpoints of the domain or at these critical numbers. x f(x) From this table, we can see the maximum activity level of the lizard occurs approximately hours after 12 noon and the minimum activity level of the lizard occurs at approximately hours after 12 noon. These times are 5:04:32 PM and 6:55:68 AM, respectively. A-4

8 18. A company wishes to run a utility cable from point A on the shore to an installation point B on the island. The island is 6 miles from the shore. Point A is 9 miles from Point C, the point on the shore is closest to Point B. To run the cable underwater, it costs 5/4 the price of running the cable on land (per mile). Assume that the cable starts at A and runs along the shoreline, then angles and runs underwater to the island. Find the point at which the line should begin to angle in order to yield the minimum total cost. (a) The hypotenuse of the triangle is x 2 and y = 9 x (b) 0 x 9 (c) There a few options here because we are not given specific costs. Here are two (I will be using the first for the rest of my answers): C(x) = x 2 + 4(9 x) or C(x) = x 2 + (9 x). (d) C (x) = 5x 6 2 +x 2 4 (e) Critical numbers: x = ± = ±8. But, 8 can be eliminated since it is not in the domain. (f) x C(x) If you used a different function in (c) your numbers be be different. The ratio between the C(x) values should be the same on your table as they are on mine. (g) The minimum cost is achieved when the cable travels 1 mile before angling toward the island. The cost is 54 $ mi if it cost $5 per mile to run the cable underwater and $4 per mile on land. Remark. If you found the real cost per mile on land to be n then you would simply multiply $54 by n 4 to get the actual minimum cost. For example, if it cost $400 per mile to run the cable on land, then the minimum cost for this problem would still occur when the cable travels 1 mile before angling toward the island and would be $ = $5400. A-5

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