Questions and Answers #13 Analyzing Functions

Questions and Answers Sheet 13 Analyzing Functions Question #1 𝑓(𝑥) = 10𝑥 5 − 7𝑥 4 + 𝑥 − 4 concave or convex at 𝑥 = −1 ? Answer: Step 1 We will find the second derivative and substitute the given value of x; if 𝑓 ′′ (−1) > 0 then it is convex: 𝑓 ′ (𝑥) = 50𝑥 4 − 28𝑥 3 + 1 𝑓 ′′ (𝑥) = 200𝑥 3 − 84𝑥 2 𝑓 ′′ (−1) = −200 − 84 < 0 Then f is concave at 𝑥 = −1 Question #2 𝑓(𝑥) = 4𝑥 5 − 12𝑥 4 + 5𝑥 2 + 2𝑥 + 2 concave or convex at 𝑥 = −5 ? Answer: Step 1 𝑓(𝑥) = 4𝑥 5 − 12𝑥 4 + 5𝑥 2 + 2𝑥 + 2 𝑓 ′ (𝑥) = 20𝑥 4 − 48𝑥 3 + 10𝑥 + 2 𝑓 ′′ (𝑥) = 80𝑥 3 − 144𝑥 2 + 10 𝑓 ′′ (−5) = 80(−5)3 − 144(−5)2 + 10 𝑓 ′′ (−5) = −10000 − 3600 + 10 = −13590 Definitely negative, so it is concave. Question #3 𝑓(𝑥) = 4𝑥 5 + 2𝑥 3 − 2𝑥 2 + 2𝑥 + 8 concave or convex at 𝑥 = −3 ? Answer: Step 1 To determine if a function is concave/convex at f(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 hence 𝑓(𝑥) = 4𝑥 5 + 2𝑥 3 − 2𝑥 2 + 2𝑥 + 8 ⇒ 𝑓 ′ (𝑥) = 20𝑥 4 + 6𝑥 2 − 4𝑥 + 2 and 𝑓 ′′ (𝑥) = 80𝑥 3 + 12𝑥 − 4 ⇒ 𝑓 ′′ (−3) = 80(−3)3 + 12(−3) − 4 = −2200 Since 𝑓 ′′ (−3) < 0 , then f(x) is concave at 𝑥 = −3 Question #4 The function: 𝑓(𝑥) = −𝑥 5 − 5𝑥 3 − 3𝑥 2 + 12𝑥 + 8 concave or convex at 𝑥 = −3 ? Answer: Step 1 We calculate the first and second derivative 𝑓(𝑥) = −𝑥 5 − 5𝑥 3 − 3𝑥 2 + 12𝑥 + 8 𝑓 ′ (𝑥) = −5𝑥 4 − 15𝑥 2 − 6𝑥 + 12 𝑓 ′′ (𝑥) = −20𝑥 3 − 30𝑥 − 6 And 𝑓(−3) = −20 ⋅ (−3)3 − 30 ⋅ (−3) − 6 = −20 ⋅ (−27) + 90 − 6 = 624 Therefore, As 𝑓 ′′ (−3) > 0, the curve is convex at 𝑥 = −3 Question #5 To calculate: 𝑓(𝑥) = −3𝑥 5 − 2𝑥 4 − 6𝑥 3 + 𝑥 − 7concave or convex at 𝑥 = 5 ? Answer: Step 1 It's a pretty straightforward polynomial, so taking the derivative is pretty easy. You can just derive the terms one by one, proceeding left to right: 𝑑 𝑑𝑥 − 3𝑥 5 − 2𝑥 4 − 6𝑥 3 + 𝑥 − 7 = −15𝑥 4 − 8𝑥 3 − 18𝑥 2 + 1 And then, do it again: 𝑑 − 15𝑥 4 − 8𝑥 3 − 18𝑥 2 + 1 𝑑𝑥 = −60𝑥 3 − 24𝑥 2 − 36𝑥 You can plug in 𝑥 = 5 and calculate if you want, but if you're pressed for time on an exam it's not really necessary. Each term in the second derivative is negative for 𝑥 = 5 , so the second derivative is therefore negative at that point, so it's convex (or concave downward) at 𝑥 = 5 . Question #6 To determine: 𝑓(𝑥) = −𝑥 5 − 2𝑥 4 + 8𝑥 2 − 4𝑥 + 2 concave or convex at 𝑥 = −5 ? Answer: Step 1 To test if a function is concave / convex at f(a) , 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 hence 𝑓(𝑥) = −𝑥 5 − 2𝑥 4 + 8𝑥 2 − 4𝑥 + 2 𝑓 ′ (𝑥) = −5𝑥 4 − 8𝑥 3 + 16𝑥 − 4 and 𝑓 ′′ (𝑥) = −20𝑥 3 − 24𝑥 2 + 16 ⇒ 𝑓(−5) − −20(−5)3 − 24(−5)2 + 16 = 1916 since 𝑓 ′′ (−5) > 0 then f(x) is convec at 𝑥 = −5 Question #7 What are the inflection points for 𝑓(𝑥) = −𝑥 4 − 9𝑥 3 + 2𝑥 + 4? Answer: Explanation: 𝑓 ′ (𝑥) = −4𝑥 3 − 27𝑥 2 + 2 𝑓=-12x^{2 − 54𝑥 Inflection points would be given by 𝑓 ′′ (𝑥) = 0 , that is −12𝑥 2 − 54𝑥 = 0 = −6𝑥(2𝑥 + 9) = 0 9 𝑥 = 0, − 2 are the inflection points. Question #8 What intervals is 𝑓(𝑥) = 5𝑥 2 𝑥 2 +4 concave up/down? Answer: Step 1 In order to investigate concavity, we shall look at the sign of the second derivative. 5𝑥 2 𝑥2 + 4 𝑓(𝑥) = −40𝑥 𝑓 ′ (𝑥) = (𝑥 2 +4)2 Finding 𝑓"(𝑥): 2 𝑓’’(𝑥) = = = −40(𝑥 2 +4) −(−40𝑥)2(𝑥 2 +4)(2𝑥) (𝑥 2 +4)4 (𝑥 2 + 4)[−40(𝑥 2 + 4) + 160𝑥 2 ] (𝑥 2 + 4)4 −40(𝑥 2 +4)+160𝑥 2 (𝑥 2 +4)3 Step 2 120𝑥 2 − 160 (𝑥 2 + 4)3 𝑓’’(𝑥) = Step 2 In general, a function can change sign by either crossing the x axis (being equal to 0), or by being discontinuous (teleporting across the x axis). In this case f"(x) is never discontinuous for real x, and 𝑓(x)=0 \text{when3𝑥 2 − 4 = 0, so𝑥 = ± 2 √3 Cut the number line into intervals and test each interval: (−∞, − (− ( 2 , 2 √3 2 √3 √3 2 √3 ) 𝑓’’(𝑥) is positive, so the graph of f is concave up ) 𝑓’’(𝑥) is negative, so the graph of f is concave down , ∞) 𝑓’’(𝑥) is positive, so the graph of f is concave up