Differential Calculus.Uses of the Derivative

Differential Calculus: Uses of the Derivative July 15, 2014 Increasing and Decreasing Functions Question 1: Which of the following functions, when defined on the set of non-negative real numbers, is increasing? (A) f (x) = 13 − x3 (B) f (x) = −4x + 1 (C) f (x) = −4x2 (D) x2 (E) None of the above Answer: (D) A function is increasing over a given domain if and only if the function has a positive first derivative when evaluated at every point within that domain. In this case the domain is the set of positive real numbers. Taking the derivatives of the above functions yields: (A) f 0 (x) = −3x2 , (B) f 0 (x) = −4, (C) f 0 (x) = −8x, (D) f 0 (x) = 2x. Of these derivatives, only (D) f 0 (x) = 2x is positive when defined over the positive real numbers. Question 2: Which of the following functions, when defined on the set of non-negative real numbers, is decreasing? (A) f (x) = 2x 1 (B) f (x) = x 2 (C) x3 + 4x2 + x + 1 (D) −2(x2 + 9) (E) None of the above Answer: (D) A function is decreasing over a given domain if and only if the function has a negative first derivative when evaluated at every point within that domain. Taking first derivatives yields: (A) f 0 (x) = (ln 2)2x , (B) f 0 (x) = 2√1 x , (C) f 0 (x) = 3x2 + 8x + 1, (D) f 0 (x) = −4x. Of these derivatives, only (D) f 0 (x) = −4x < 0 for all positive real numbers x. Question 3: Is f (x) increasing or decreasing at point x = 3? f (x) = 5x2 − 4x − 89 Answer: Take the derivative and evaluate it at point x = 3. f 0 (x) = 10x − 4 f 0 (3) = 10(3) − 4 = 26 > 0 As f 0 (3) > 0, the function is increasing at that point. Question 4: Is f (x) increasing or decreasing at point x = 2? f (x) = 2x3 − 17x2 + 5x + 127 1 Answer: Take the derivative and evaluate it at point x = 2. f 0 (x) = 6x2 − 34x + 5 f 0 (2) = 6(2)2 − 34(2) + 5 = −39 < 0 As f 0 (2) < 0, the function is decreasing at that point. Concavity, and Convexity Question 5: Which of the following functions is strictly concave over the set of non-negative real numbers? (A) f (x) = 2x 1 (B) f (x) = x 2 (C) f (x) = 3x + 2 (D) f (x) = x3 (E) None of the above Answer: (B) A function f (x) is strictly concave at the point x = a if and only if f 00 (a) < 0. The sign of the first derivative is irrelevant in determining whether or not the function is concave! Check second derivatives. (A) f 00 (x) = (ln 2)2 2x , (B) f 00 (x) = − √1 3 , (C) f 00 (x) = 0, f 00 (x) = 6x. We are looking for a function 4 x that is strictly concave over the entire set of non-negative real numbers. Only (B) has a negative second derivative for all non-negative real numbers. Question 6: Which of the following functions is strictly convex over the set of non-negative real numbers? (A) f (x) = 2x 1 (B) f (x) = x 2 (C) f (x) = 3x + 2 (D) f (x) = x3 − x2 + x (E) None of the above Answer: (A) A function f (x) is strictly convex at the point x = a if and only if and f 00 (a) > 0. The sign of the first derivative is irrelevant in determining whether or not the function is convex! Check second derivatives. (A) f 00 (x) = (ln 2)2 2x , (B) f 00 (x) = − √1 3 , (C) f 00 (x) = 0, (D) f 00 (x) = 6x − 2. We are looking for a 4 x function that is strictly convex over the entire set of non-negative real numbers. Only (A) has a positive second derivative over the entire set of real numbers (f 00 (x) = 6x − 2 > 0 if and only if x > 13 and thus f (x) = x3 − x2 + x is not convex over the set of non-negative real numbers). Question 7: Describe the concavity of convexity of f (x) over its domain 0 ⇔ (4 − 9x) > 0 ⇔ (1) 4 >x 9 As f 00 (x) > 0∀x < 94 , the function is convex for those values, and concave ∀x > 49 . 2 (2) Question 8: Describe the concavity of convexity of f (x) over its domain 0 ⇔ 30x − 162 > 0 ⇔ x > 5.4 (3) (4) As f 00 (x) > 0∀x > 5.4, the function is convex for those values, and concave ∀x < 5.4. Relative Extrema Question 9: Find the relative extrema for the following functions by (1) determining the critical value(s) and (2) determining whether at these critical values the function is a relative maximum or minimum (or possible inflection point). f (x) = −8x2 + 12x + 3 (A) x = 34 , relative maximum (B) x = 34 , relative minimum (C) x = − 34 , relative minimum (D) x = − 43 , relative maximum (E) None of the above Answer: (A) To find the critical values of the function, take the first derivative of the function, set it equal to zero and solve for x: f 0 (x) = −16x + 12 = 0 implies that x = 34 . Now need to determine whether or not the function is concave or convex at x = 43 (recall that concavity implies relative maximum, convexity implies relative minimum!). f 00 (x) = −16 < 0 implies that f (x) is concave. Thus the function f (x) attains a relative maximum at the point x = 43 . Question 10: Find the relative extrema for the following functions by (1) determining the critical value(s) and (2) determining whether at these critical values the function is a relative maximum or minimum (or possible inflection point). f (x) = 2x3 + 18x2 + 36x + 20 √ √ (A) x = −3 − 2 3, relative maximum; x = −3 + 2 3, relative minimum (B) x = −3 relative maximum √ √ (C) x = 3 − 3, relative maximum; x = 3 + 3, relative minimum √ √ (D) x = −1 − 3, relative maximum; x = −1 + 3, relative minimum (E) None of the above Answer: (E) To find the critical values of the function, take the first derivative of the function, set it equal to zero and solve for x: f 0 (x) = 6x2 + 36x + 36 = 0 which implies that x2 + 6x + 6 = 0. Using quadratic formula 3 yields critical values of: p 62 − (4)(6) √ 2 −6 ± 36 − 24 = 2 √ −6 ± 2 3 = 2√ = −3 ± 3 x= −6 ± √ Now need to determine whether or not the function is concave or convex at x = −3 ± 3 (again recall that concavity implies relative maximum, convexity√implies relative√minimum!). √ Second derivative of the function √ is 00 00 f (x) = 12x + √36 which implies √ that f (−3 √+ 3) = 12(−3 + 3) + 36 = 3√> 0 (f (x) convex at x = −3 + 3) and f 00 (−3 − 3) = 12(−3 − 3) + 36 = − √ 3 < 0 (f (x) concave at x = −3 − 3). Thus √ the function f (x) attains a relative maximum at the point x = −3 − 3 and a relative minimum at x = −3 + 3. Question 11: Find the relative extrema for the following functions by (1) determining the critical value(s) and (2) determining whether at these critical values the function is a relative maximum or minimum (or possible inflection point). f (x) = (x − 1)4 (A) x = 1, inflection point (B) x = 1, relative minimum (C) x = 1, relative maximum (D) x = 1, inconclusive (E) None of the above Answer: (D) To find the critical values need to take the first derivative and set it equal to zero: f 0 (x) = 4(x − 1)3 = 0 implies that x = 1 is the only critical point! Check second derivative to determine concavity or convexity: f 00 (x) = 12(x − 1)2 thus f 00 (1) = 0! Test is inconclusive. Need to apply the successive-derivative test by checking higher order derivatives: f 000 (x) = 24(x − 1) and f 000 (1) = 0 (still inconclusive!); f 0000 (x) = 24 > 0 (since the first nonzero higher order derivative encountered is an even numbered derivative (i.e., the fourth derivative!) we know that the function attains a relative extremum at the point x = 1; because f 0000 (1) = 24 > 0 the function is convex at x = 1 and therefore the function attains a relative minimum at x = 1. Question 12: Find the relative extrema and inflection points of f (x). f (x) = −5x3 + 135x2 − 675x + 107 Answer: First take the first derivative and set it equal to zero: f 0 (x) = −15x2 + 270x − 675 f 0 (x) = 0 x = 3 ∨ x = 15 (5) Now take the second derivative and substitute the two critical values for x. f 00 (x) = −30x + 270 f 00 (3) = 180 > 0 f 00 (15) = −180 > 0 (6) 4 Therefore 3 is a minimum and 15 is a maximum. Finally, solve f 00 (x) = 0 to determine the existence of any inflection points. f 00 (x) = 0 −30x + 270 = 0 x=9 (7) As x = 9 is not an extremum, it is an inflection point. Question 13: Find the relative extrema and inflection points of f (x). f (x) = (x + 6)3 Answer: First take the first derivative and set it equal to zero: f 0 (x) = 3(x + 6)2 f 0 (x) = 0 x = −6 (8) Now take the second derivative and substitute the critical value for x. f 00 (x) = 6(x + 6) f 00 (−6) = 0 As the second derivative is equal to 0 at the critical point, we say anything about the point. Perform the higher derivative test: f 000 (x) = 6 > 0 The third derivative is the first non-zero higher derivative. As that is an odd numbered derivative, −6 is an inflection point and the function has no maxima or minima. Question 14: Find the relative extrema and inflection points of f (x). f (x) = −3(4x − 11)6 Answer: First take the first derivative and set it equal to zero: f 0 (x) = −72(4x − 11)5 f 0 (x) = 0 11 x= 4 (9) Now take the second derivative and substitute the critical value for x. f 00 (x) = −1440(4x − 11)4 11 f 00 ( ) = 0 4 As the second derivative is equal to 0 at the critical point, we say anything about the point. Perform the higher derivative test: f 000 (x) = −23040(4x − 11)3 f 000 ( 11 11 11 ) = 0f 0000 (x) = −276480(4x − 11)2 f 0000 ( ) = 0f 00000 (x) = −2211840(4x − 11)f 00000 ( ) 4 4 4 The sixth derivative is the first non-zero higher derivative. As that is an even numbered derivative, −6 is an extremum. As the value of the sixth derivative is negative, it is a maximum. 5 Optimization Question 15: Optimize the following function by (1) finding the critical value(s) at which the function is optimized and (2) testing the second-order condition to distinguish between a relative maximum or minimum and (3) the values of the relative extrema for the function. f (x) = x2 + 6x + 9 (A) x = −3, relative maximum, f (−3) = 0 (B) x = 3, relative minimum, f (−3) = 0 (C) x = −3, relative minimum, f (−3) = 0 (D) x = −3, relative minimum, f (−3) = −1 (E) None of the above Answer: (C) To find the critical points, take the first derivative and set it equal to zero: f 0 (x) = 2x + 6 = 0 implies that there is a single critical point at x = −3. Evaluate second derivative at x = −3 to check for concavity or convexity of function at that point: f 00 (x) = 2. Thus f 00 (−3) = 2 > 0 and the function is convex at x = −3 (relative minimum). Thus the function obtains its minimum value of f (−3) = 0 at x = −3. Question 16: Optimize the following function by (1) finding the critical value(s) at which the function is optimized and (2) testing the second-order condition to distinguish between a relative maximum or minimum and (3) the value(s) of the relative extrema for the function. f (x) = x3 + 6x2 − 96x + 23 (A) x = 4, relative minimum; x = −8, relative maximum; f (4) = −201 and f (−8) = 663 (B) x = −4, relative minimum; x = 8, relative maximum; f (−4) = 200 and f (8) = −665 (C) x = 4, relative minimum; x = 8, relative maximum; f (4) = −201 and f (8) = 663 (D) x = 4, relative minimum; x = −8, relative maximum; f (4) = −154 and f (−8) = 653 (E) None of the above Answer: (A) To find critical values take first derivative and set it equal to zero, then solve for x: f 0 (x) = 3x2 + 12x − 96 = 0 implies that 3(x2 + 4x − 32) = 0 which implies that (x − 4)(x + 8) = 0. Thus we have critical values of x = 4 and x = −8. Check the sign of the second derivative to determine concavity or convexity: f 00 (x) = 6x + 12, thus f 00 (4) = 36 > 0 which implies f (x) is convex at x = 4 (relative minimum); f 00 (−8) = −36 < 0 which implies that f (x) is concave at x = −8 (relative maximum). Relative extrema of the function are f (4) = 43 + 6(42 ) − 96(4) + 23 = −201 and f (−8) = (−8)3 + 6(−8)2 − 96(−8) + 23 = 663. Question 17: Optimize the following function by (1) finding the critical value(s) at which the function is optimized and (2) testing the second-order condition to distinguish between a relative maximum or minimum and (3) the value(s) of the relative extrema for the function. f (x) = −4x3 + 186x2 − 1008x + 25 Answer: Take the first derivative and set it equal to zero, then solve for x: f 0 (x) = −12x2 + 372x − 1008 f 0 (x) = 0 x = 3 ∨ x = 28 (10) 6 Now take the second derivative and substitute the critical values for x. f 00 (x) = −24x + 372 f 00 (3) = 300 > 0 f 00 (28) = −300 < 0 As f 00 (3) > 0, 3 is a minimum; as f 00 (28) < 0, 28 is a maximum. Question 18: Optimize the following function by (1) finding the critical value(s) at which the function is optimized and (2) testing the second-order condition to distinguish between a relative maximum or minimum and (3) the value(s) of the relative extrema for the function. f (x) = 4x4 − 48x3 − 288x2 + 129 Answer: Take the first derivative and set it equal to zero, then solve for x: f 0 (x) = 16x3 − 144x2 − 576x = x(16x2 − 144x − 576) f 0 (x) = 0 x = −3 ∨ x = 0 ∨ x = 12 (11) Now take the second derivative and substitute the critical values for x. f 00 (x) = 48x2 − 288x − 576 f 00 (−3) = 720 > 0 f 00 (0) = −576 < 0 f 00 (12) = 2880 > 0 (12) As f 00 (−3) > 0, −3 is a minimum; as f 00 (0) < 0, 0 is a maximum; as f 00 (12) > 0, 12 is a minimum. Question 19: Optimize the following function by (1) finding the critical value(s) at which the function is optimized and (2) testing the second-order condition to distinguish between a relative maximum or minimum and (3) the value(s) of the relative extrema for the function. f (x) = x4 + 36x3 + 280x2 − 79 Answer: Take the first derivative and set it equal to zero, then solve for x: f 0 (x) = 4x3 + 108x2 + 560x = x(4x2 + 108x + 560) f 0 (x) = 0 x = −7 ∨ x = 0 ∨ x = −20 (13) Now take the second derivative and substitute the critical values for x. f 00 (x) = 12x2 + 216x + 560 f 00 (−20) = 1040 > 0 f 00 (−7) = −364 < 0 f 00 (0) = 560 > 0 As f 00 (−20) > 0, −20 is a minimum; as f 00 (−7) < 0, −7 is a maximum; as f 00 (0) > 0, 0 is a minimum. 7 (14)