Questions and Answers #12 Analyzing Functions

Questions and Answers Sheet 12 Analyzing Functions Question #1 Is 𝑓(𝑥) = 1 − 𝑥𝑒 −3𝑥 concave or convex at 𝑥 = −2 Answer: Answer: function is convex at 𝑥 = −2 Step 1 Given function: 𝑓(𝑥) = 1 − 𝑥𝑒 −3𝑥 𝑓 ′ (𝑥) = −𝑥(−3𝑒 −3𝑥 ) − 𝑒 −3𝑥 (1) 𝑓 ′ (𝑥) = 𝑒 −3𝑥 (3𝑥 − 1) 𝑓(x)=e^{-3x(3) + (3𝑥 − 1)(−3𝑒 −3𝑥 ) Step 2 𝑓(x)=e^{-3x(6 − 9𝑥) 𝑓(-2)=e^{6(6 − 9(−2)) = 24𝑒 6 Since f(-2)>0 hence the given function is convex at 𝑥 = −2 Question #2 Is 𝑓(𝑥) = 3𝑥 5 − 𝑥 3 + 8𝑥 2 − 𝑥 + 3 concave or convex at 𝑥 = −4 ? Answer: Answer: To find the concavity of a function f(x), you will need to determine the second derivative, and then evaluate at at the given point. If the result of 𝑓’‘(𝑎) > 0 , the function is concave up; if the result is negative, it it concave down. 𝑓(𝑥) = 3𝑥 5 − 𝑥 3 + 8𝑥 2 − 𝑥 + 3 𝑓 ′ (𝑥) = 15𝑥 4 − 3𝑥 2 + 16𝑥 = 1 𝑓 ′′ (𝑥) = 60𝑥 3 − 6𝑥 + 16 𝑓 ′′ (−4) = 60(−4)3 − 6(−4) + 16 = −3,800 < 0 Thus, f(x) is concave down at 𝑥 = −4 . Question #3 Is 𝑓(𝑥) = 3𝑥 5 − 𝑥 3 + 8𝑥 2 − 𝑥 concave or convex at 𝑥 = 0 ? Answer: Step 1 f(x) is concave at 𝑥 = 𝑎 if 𝑓’‘(𝑎) > 0 . f(x) is convex at 𝑥 = 𝑎 if 𝑓’‘(𝑎) > 0 . Here, 𝑓(𝑥) = 3𝑥 5 − 𝑥 3 + 8𝑥 2 − 𝑥, 𝑎 = 0 . Knowing that, let's take the second derivative. 𝑓 ′ (𝑥) = 15𝑥 4 − 3𝑥 2 + 16𝑥 − 1 𝑓’‘(𝑥) = 20𝑥 3 − 6𝑥 + 16 Step 2 Evaluate at 0: 𝑓’‘(0) = 20(03 ) − 6(0) + 16 = 16 > 0 f(x) is concave at 𝑥 = 0 . Question #4 Is 𝑓(𝑥) = 4𝑥 5 − 2𝑥 4 − 9𝑥 3 − 2𝑥 2 − 6𝑥 concave or convex at 𝑥 = −1 ? Answer: Step 1 Concavity and convexity are determined by the sign of the second derivative: - If 𝑓’‘(−1) < 0 , then the function is concave at 𝑥 = −1 . - If 𝑓’‘(−1) > 0 , then the function is convex at 𝑥 = −1 . Step 2 Find the second derivative: 𝑓(𝑥) = 4𝑥 5 − 2𝑥 4 − 9𝑥 3 − 2𝑥 2 − 6𝑥 𝑓 ′ (𝑥) = 20𝑥 4 − 8𝑥 3 − 27𝑥 2 − 4𝑥 − 6 𝑓 ′′ (𝑥) = 80𝑥 3 − 24𝑥 2 − 54𝑥 − 4 Find the sign of the second derivative when 𝑥 = −1 : 𝑓 ′′ (−1) = 80(−1)3 − 24(−1)2 − 54(−1) − 4 = 80(−1) − 24(1) + 54 − 4 = −80 − 24 + 50 = −54 Since 𝑓’‘(−1) < 0 , the function is concave at 𝑥 = −1 . This means that it will resemble the ∩ shape. Question #5 Is 𝑓(𝑥) = 4𝑥 5 − 𝑥 4 − 9𝑥 3 + 5𝑥 2 − 7𝑥 concave or convex at 𝑥 = −1 ? Answer: Step 1 To determine if a function is concave/convex we calculate the value of 𝑓 ′′ (𝑥) 𝑎𝑡 𝑥 = 𝑎 . - If 𝑓’‘(𝑎) > 0, then f(x) is convex at 𝑥 = 𝑎 - If 𝑓’‘(𝑎) < 0 , then f(x) is concave at 𝑥 = 𝑎 𝑓(𝑥) = 4𝑥 5 − 𝑥 4 − 9𝑥 3 + 5𝑥 2 − 7𝑥 Step 2 Differentiate using the power rule ⇒ 𝑓 ′ (𝑥) = 20𝑥 4 − 4𝑥 3 − 27𝑥 2 + 10𝑥 − 7 and 𝑓 ′′ (𝑥) = 80𝑥 3 − 12𝑥 2 − 54𝑥 + 10 so 𝑓’‘(−1) = 80(−1)3 − 12(−1)2 − 54(−1) + 10 = −80 − 12 + 54 + 10 = −28 Question #6 Is 𝑓(𝑥) = 5𝑥 5 − 𝑥 2 + 4𝑥 + 3 concave or convex at 𝑥 = −4 ? Answer: Step 1 Calculate the first and second derivatives 𝑓(𝑥) = 5𝑥 5 − 𝑥 2 + 4𝑥 + 3 𝑓 ′ (𝑥) = 25𝑥 4 − 2𝑥 + 4 𝑓 ′′ (𝑥) = 100𝑥 3 − 2 When 𝑥 = −4 𝑓 ′′ (−4) = 100 ⋅ (−4)3 − 2 = −6402 As 𝑓 ′′ (𝑥) < 0 , the function is concave at 𝑥 = −4 Question #7 Is 𝑓(𝑥) = 𝑒 𝑥−1 − 𝑥2 𝑥−1 concave or convex at 𝑥 = −1 ? Answer: Step 1 The derivative of a quotient is 𝑢 ′ 𝑢′ 𝑣 − 𝑢𝑣 ′ ( ) = 𝑣 𝑣2 The first derivative is 𝑓(𝑥) = 𝑒 𝑥−1 − 𝑥2 −1 𝑥−1 𝑓 ′ (𝑥) = 𝑒 𝑥−1 − (2𝑥(𝑥 − 1) − 𝑥 2 ) (𝑥 − 1)2 𝑓 ′ (𝑥) = 𝑒 𝑥−1 − 𝑥 2 − 2𝑥 (𝑥 − 1)2 Step 2 The second derivative is 𝑓’‘(𝑥) = 𝑒 𝑥−1 − (2𝑥 − 2)(𝑥 − 1)2 − 2(𝑥 2 − 2𝑥)(𝑥 − 1) (𝑥 − 1)4 = 𝑒 𝑥−1 − 2𝑥 2 − 2𝑥 − 2𝑥 + 2 − 2𝑥 2 + 4𝑥 (𝑥 − 1)3 = 𝑒 𝑥−1 − 2 (𝑥 − 1)3 At 𝑥 = −1 2 𝑓’‘(−1) = 𝑒 −1−1 − (−1−1)3 = 0.135 + 0.25 = 0.385 As 𝑓’‘(−1) > 0 , we conclude that f(x) is convex at 𝑥 = −1 Question #8 Is 𝑓(𝑥) = 𝑒 𝑥 sin 𝑥 − cos 𝑥 concave or convex at 𝑥 = 𝜋 3 ? Answer: Step 1 Obtain f'(x) by differentiating the original function (product rule): 𝑓 ′ (𝑥) = 𝑒 𝑥 sin(𝑥) + 𝑒 𝑥 cos(𝑥) + sin(𝑥) Obtain f’‘(x) by differentiating again: 𝑓 ′′ (𝑥) = 𝑒 𝑥 𝑠𝑖𝑛(𝑥) + 𝑒 𝑥 𝑐𝑜𝑠(𝑥) + 𝑒 𝑥 𝑐𝑜𝑠(𝑥) − 𝑒 𝑥 𝑠𝑖𝑛(𝑥) + 𝑐𝑜𝑠(𝑥) = 2𝑒 𝑥 𝑐𝑜𝑠(𝑥) + 𝑐𝑜𝑠(𝑥) = (2𝑒 𝑥 + 1) cos(𝑥) Step 2 𝜋 Evaluate 𝑓’‘ ( ): 3 𝜋 𝜋 𝜋 𝑓’‘ ( ) = (2𝑒 3 + 1) cos ( ) 3 3 𝜋 1 = (2𝑒 3 + 1) 2 𝜋 𝜋 3 3 Clearly 𝑓’‘ ( ) > 0 , so the function will be concave up at 𝑥 = Note that the naming convention I learnt was concave up and concave down, so I am not sure which is concave or convex in your terms. Graphing the original fucntion is a good way to check your answer, at 𝑥 = change of gradient is indeed positive and therefore concave up. Question #9 Is 𝑓(𝑥) = 𝑒 𝑥 − 𝑥2 𝑥−1 concave or convex at 𝑥 = 0 ? Answer: Step 1 The derivative of a quotient is 𝑢 𝑢′ 𝑣−𝑢𝑣 ′ 𝑣 𝑣2 ( )= The function is 𝑓(𝑥) = 𝑒 𝑥 − 𝑥2 𝑥−1 The first derivative is 𝑓 ′ (𝑥) = 𝑒 𝑥 − 2𝑥(𝑥−1)−𝑥 2 ⋅1 (𝑥−1)2 𝑥 2 −2𝑥 = 𝑒 𝑥 − (𝑥−1)2 Step 2 The second derivative is 𝑓’‘(𝑥) = 𝑒 𝑥 − = 𝑒𝑥 − (2𝑥 − 2)(𝑥 − 1)2 − 2(𝑥 2 − 2𝑥)(𝑥 − 1) (𝑥 − 1)2 (2𝑥 − 2)(𝑥 − 1) − (2𝑥 2 − 4𝑥) (𝑥 − 1)3 2𝑥 2 − 4𝑥 + 2 − 2𝑥 2 + 4𝑥 (𝑥 − 1)3 2 = 𝑒𝑥 − (𝑥 − 1)3 = 𝑒𝑥 − Step 3 When 𝑥 = 0 𝜋 3 ≈ 1 , we see that the rate of 2 𝑓’‘(0) = 𝑒 0 − (−1)3 = 1 − 2 = 3 As 𝑓’‘(0) > 0 , the function is convex at 𝑥 = 0 . Question #10 Is 𝑓(𝑥) = 𝑥 cos 𝑥 concave or convex at 𝑥 = 𝜋 2 ? Answer: Step 1 Given: 𝑓(𝑥) = 𝑥 cos 𝑥 𝑓 ′ (𝑥) = 𝑥(− sin 𝑥) + cos 𝑥 = −𝑥 sin 𝑥 + cos 𝑥 𝑓’‘(𝑥) = −𝑥(cos 𝑥) − sin 𝑥 − sin 𝑥 𝑓’‘(𝑥) = −𝑥 cos 𝑥 − 2 sin 𝑥 Step 2 𝜋 𝜋 𝜋 𝜋 𝑓’‘ ( ) = − cos ( ) − 2 sin ( ) 2 2 2 2 = 0 − 2(1) = −2 𝜋 𝜋 2 2 Since 𝑓 ( ) < 0 hence the given function is concave at 𝑥 = Question #11 Is 𝑓(𝑥) = 𝑥 ln 𝑥 − 𝑥 concave or convex at 𝑥 = 1 ? Answer: Step 1 f(x) is concave upward (convex) at a point 𝑥0 if 𝑓’‘(𝑥0 ) > 0 and concave downward (concave) at a point 𝑥0 if 𝑓’‘(𝑥0 ) < 0 . We have 𝑓’‘(𝑥) = 𝑑 𝑑𝑥 𝑓 ′ (𝑥) = 𝑑 𝑑 ( 𝑥 ln(𝑥) − 𝑥) 𝑑𝑥 𝑑𝑥 = 𝑑 ln(𝑥) 𝑑𝑥 = 1 𝑥 Step 2 Then, at 𝑥 = 1 : 𝑓’‘′ (1) = 1 =1>0 1 Thus f(x) is convex at 𝑥 = 1 . Question #12 Is 𝑓(𝑥) = 𝑥(𝑥 − 2)(𝑥 + 3)(𝑥 − 1) concave or convex at 𝑥 = 1 ? Answer: Step 1 This problem may be easier if we determine the expanded (not factored) form of f(x) first. 𝑓(𝑥) = 𝑥(𝑥 − 2)(𝑥 + 3)(𝑥 − 1) = 𝑥(𝑥 2 + 𝑥 − 6)(𝑥 − 1) = 𝑥(𝑥 3 − 7𝑥 + 6) = 𝑥 4 − 7𝑥 2 + 6𝑥 = 𝑥 4 − 7𝑥 2 + 6𝑥 𝑓 ′ (𝑥) = 4𝑥 3 − 14𝑥 + 6 Step 2 and 𝑓 ′′ (𝑥) = 12𝑥 2 − 14 𝑓 ′′ (1) = 12 ⋅ 12 − 14 = −2 Which means that f(x) is concave DOWNWARDS (convex?) at 𝑥 = 1 because 𝑓’‘(1) < 0 Question #13 Is 𝑓(𝑥) = 𝑥 3 − 2𝑥 + 7 concave or convex at 𝑥 = 2 ? Answer: Step 1 To determine if a function f(x) is concave/convex at 𝑥 = 𝑎 we require to find the value of f''(a). - If 𝑓 ′′ (𝑎) > 0 then f(x) is convex at 𝑥 = 𝑎 - If 𝑓 ′′ (𝑎) < 0 then f(x) is concave at 𝑥 = 𝑎 Step 2 Now 𝑓(𝑥) = 𝑥 3 − 2𝑥 + 7 differentiate using the power rule ⇒ 𝑓 ′ (𝑥) = 3𝑥 2 − 2 and 𝑓’‘(𝑥) = 6𝑥 ⇒ 𝑓’‘(2) = (6 × 2) = 12 Since 𝑓’‘(2) > 0 then f(x) is convex at 𝑥 = 2 . Question #14 2 Is 𝑓(𝑥) = 𝑥 3 − 𝑥 + 𝑒 𝑥−𝑥 concave or convex at 𝑥 = 1 ? Answer: Explanation: The function will be convex if the second derivative is positive, and concave if it is negative. Step 1 The first derivative is: 3𝑥 2 + (−2𝑥 + 1)𝑒 −𝑥 2 +𝑥 The second derivative is: 6𝑥 + (2𝑥 − 1)2 𝑒 −𝑥 −1 2 +𝑥 − 2𝑒 −𝑥 2 +𝑥 Step 2 At 𝑥 = 1 : 6 + 2𝑒 2 = 20.78 positive value Question #15 Is 𝑓(𝑥) = 𝑥 4 − 2𝑥 3 − 9𝑥 − 14 concave or convex at 𝑥 = −1? Answer: Step 1 4 Given: 𝑓(𝑥) = 𝑥 𝑥 − 2𝑥 3 − 9𝑥 − 14 Find the derivative 𝑓 ′ (𝑥) = 4𝑥 3 − 6𝑥 2 − 9 Step 2 Differentiate again with respect to x 𝑓 2 (𝑥) = 12𝑥 2 − 12𝑥 Step 3 Substitute 𝑥 = −1 in 𝑓 2 (𝑥) and check for sign 𝑓 2 (−1) = 12(−1)2 − 12(−1) 𝑓 2 (−1) = 12 + 12 𝑓 2 (−1) = 24 If f^{2}(x)>0 then the curve is convex. If 𝑓 2 (𝑥) < 0 , then the curve is concave We can see at 𝑥 = −1 the second derivative is greater than zero, hence, the curve is convex. For further information, you can refer Note: Concave up is same as convex and concave down is concave Question #16 Is 𝑓(𝑥) = 𝑥 5 − 4𝑥 4 + 16𝑥 − 4 concave or convex at 𝑥 = −2 ? Answer: Step 1 Let's calculate the first and second derivatives 𝑓(𝑥) = 𝑥 5 − 4𝑥 4 + 16𝑥 − 4 𝑓 ′ (𝑥) = 5𝑥 4 − 16𝑥 3 + 16 𝑓 ′′ (𝑥) = 20𝑥 3 − 48𝑥 2 = 4𝑥 2 (5𝑥 − 12) Now, we calculate, 𝑓 ′′ (−2) 𝑓 ′′ (−2) = 20 ⋅ (−2)3 − 48 ⋅ (−2)2 = −160 − 192 = −352 As, 𝑓 ′′ (−2) < 0 , we conclude that f(-2) is concave ∩ Normally, we calculate the concavity over an interval for example [−∞, 0] Question #17 Is 𝑓(𝑥) = 𝑥𝑒 𝑥 − 𝑥 2 concave or convex at 𝑥 = 0 ? Answer: Step 1 You need to compute the second derivative in the point 𝑥 = 0 , and check its sign. If it's positive, the function is convex, otherwise it's concave. So, we have 𝑓(𝑥) = 𝑥𝑒 𝑥 − 𝑥 2 𝑓 ′ (𝑥) = 𝑒 𝑥 + 𝑥𝑒 𝑥 − 2𝑥 𝑓’‘(𝑥) = 𝑒 𝑥 + 𝑒 𝑥 + 𝑥𝑒 𝑥 − 2 = 𝑒 𝑥 (2 + 𝑥) − 2 So, 𝑓 ′′ (0) = 𝑒 0 (2 + 0) − 2 = 2 − 2 = 0 Step 2 Since the second derivative is negative immediatly before 0, and positive after, 𝑥 = 0 is an inflection point, in which the function passes from being concave to be convex, the scales of x and y axes are changed in order to emphatize the behaviour of the curve. Question #18 Is the function concave up or down if 𝑓(𝑥) = (ln 𝑥)2 ? Answer: Step 1 It is concave up on the interval (0, e) and concave down on (𝑒, ∞) . 𝑓(𝑥) = (ln 𝑥)2 . 𝑓 ′ (𝑥) = 2(ln 𝑥) ⋅ 1 2 ln 𝑥 = . 𝑥 𝑥 2 ( ) ⋅ 𝑥 − 2 ln 𝑥 ⋅ 1 2(1 − ln 𝑥) 𝑓’‘(𝑥) = 𝑥 = 𝑥2 𝑥2 1 − ln 𝑥 = 0 where ln 𝑥 = 1 which is at 𝑥 = 𝑒 . Step 2 All other factors of f’‘(x) are always positive, so the sign of f’‘(x) is the same as the sign of 1 − ln 𝑥 . Therefore, f’‘(x) is positive if 𝑥 < 𝑒 ( so ln 𝑥 < 1 ) and it is negative for 𝑥 > 𝑒 (where ln 𝑥 > 1 ). Question #19 On what interval is 𝑓(𝑥) = 6𝑥 3 + 54𝑥 − 9 concave up and down? Answer: Step 1 A function is concave up when the second derivative is positive, it is concave down when it is negative, and there could be an inflection point when it is zero. 𝑦 ′ = 18𝑥 2 + 54 𝑦 ′′ = 36𝑥 + 54 so: 𝑦 ′′ > 0 ⇒ 𝑥 ≻ 54 3 ⇒𝑥≻ 36 2 3 In (− , + ∞) the concave is up. 2 3 in (−∞, − ) the concave is down, 2 in 𝑥 = − 3 2 there is an inflection point. Question #20 On what intervals the following equation is concave up, concave down and where it's inflection point is (𝑥, 𝑦)𝑓(𝑥) = 𝑥 8 (ln(𝑥)) ? Answer: Step 1 To analyze concavity and inflection points of a twice differentiable function f, we can study the positivity of the second derivative. In fact, if 𝑥0 is a point in the domain of f, then: - if 𝑓’‘(𝑥0 ) > 0 , then f is concave up in a neighborhood of 𝑥0 ; - if 𝑓’‘(𝑥0 ) < 0 , then f is concave down in a neighborhood of 𝑥0 ; - if 𝑓’‘(𝑥0 ) = 0 and the sign of f'' on a sufficiently small right-neighborhood of 𝑥0 is opposite to the sign of f'' on a sufficiently small left-neighborhood of 𝑥0 , then 𝑥 = 𝑥0 is called an inflection point of f. In the specific case of 𝑓(𝑥) = 𝑥 8 ln(𝑥) . we have a function whose domain has to be restricted to the positive reals 𝑅+ . 1 The first derivative is 𝑓 ′ (𝑥) = 8𝑥 7 ln(𝑥) + 𝑥 8 = 𝑥 7 [8 ln(𝑥) + 1] 3 Step 2 8 The second derivative is 𝑓 ′′ (𝑥) = 7𝑥 6 [8 ln(𝑥) + 1] + 𝑥 7 = 𝑥 6 [56 ln(𝑥) + 15] 𝑥 Let's study the positivity of f’‘(x): −𝑥 6 > 0 ⟺ 𝑥 ≠ 0 −56 ln(𝑥) + 15 > 0 ⟺ ln(𝑥) ≻ 15 56 15 ⟺ 𝑥 > 𝑒 56 So, considering that the domain is 𝑅+ , we get that 15 - if 0 < 𝑥 < 𝑒 −56 then 𝑓’‘(𝑥) < 0 is concave down; 15 - if 𝑥 > 𝑒 −56 then 𝑓’‘(𝑥) > 0 and f is concave up; 15 - if 𝑥 = 𝑒 −56 then 𝑓’‘(𝑥) = 0 . Considering that on the left of this point f'' is negative and on the right it is 15 positive, we conclude that 𝑥 = 𝑒 −56 is a (falling) inflection point