Questions and Answers #11 Analyzing Functions

Questions and Answers Sheet 11 Analyzing Functions Question #1 Is 𝑓(𝑥) = −2𝑥 5 − 3𝑥 4 + 15𝑥 − 4 concave or convex at 𝑥 = −4? Answer: Step 1 The concavity and convexity of a function can be determined by examining the sign of a function's second derivative. Note that: you may call concave ‘‘concave down’‘ and convex ‘‘concave up.’‘ We must find the function's second derivative through the power rule: 𝑓(𝑥) = −2𝑥 5 − 3𝑥 4 + 15𝑥 − 4 𝑓 ′ (𝑥) = −10𝑥 4 − 12𝑥 3 + 15 𝑓 ′′ (𝑥) = −40𝑥 3 − 36𝑥 2 The value of the second derivative at 𝑥 = −4 is: 𝑓 ′′ (−4) = −40(−4)3 − 36(−4)2 = 1984 Since this is >0 , the function is convex (sometimes called concave up) at 𝑥 = −4 Question #2 Is 𝑓(𝑥) = −2𝑥 5 − 2𝑥 3 + 3𝑥 2 − 𝑥 + 3 concave or convex at 𝑥 = −1 ? Answer: Step 1 You can tell if a function is concave or convex by the sign of its second derivative: - If 𝑓’‘(−1) < 0 , then f(x) is concave at 𝑥 = −1 . - If 𝑓’‘(−1) > 0 , then f(x) is convex at 𝑥 = −1 . To find the second derivative, apply the power rule to each term twice. 𝑓(𝑥) = −2𝑥 5 − 2𝑥 3 + 3𝑥 2 − 𝑥 + 3 𝑓 ′ (𝑥) = −10𝑥 4 − 6𝑥 2 + 6𝑥 − 1 𝑓’‘(𝑥) = −40𝑥 3 − 12𝑥 + 6 Step 2 Find the sign of the second derivative at 𝑥 = −1 : 𝑓 ′′ (−1) = −40(−1)3 − 12(−1) + 6 This mostly becomes a test of keeping track of your positives and negatives. 𝑓 ′′ (−1) = −40(−1) + 12 + 6 = 40 + 18 = 58 Since this is >0 , the function is convex at 𝑥 = −1 . Convexity on a graph is characterized by a ∪ shape. Question #3 Is 𝑓(𝑥) = −3𝑥 5 − 2𝑥 4 − 9𝑥 2 + 𝑥 − 7 concave or convex at 𝑥 = 5 ? Answer: Step 1 𝑓 ′′ (𝑥) < 0 at 𝑥 = 5 (concave) 𝑓 ′′ (𝑥) > 0 at 𝑥 = 5 (convex) 𝑓 ′ (𝑥) = −15𝑥 4 − 8𝑥 3 − 18𝑥 + 1 𝑓 ′′ (𝑥) = −60𝑥 3 − 24𝑥 2 − 18 𝑓 ′′ (5) = −60(125) − 24(25) − 18 = −8118 Less than zero, therefore concave. Question #4 Is 𝑓(𝑥) = −8𝑥 5 − 𝑥 2 + 5𝑥 − 4 concave or convex at 𝑥 = −4? Answer: Step 1 To test if a function is concave/convex at f(a), require to find the value of f''(a) Hence 𝑓(𝑥) = −8𝑥 5 − 𝑥 2 + 5𝑥 − 4 𝑓 ′ (𝑥) = −40𝑥 4 − 2𝑥 + 5 and 𝑓 ′′ (𝑥) = −160𝑥 3 − 2 ⇒ 𝑓 ′′ (−4) = −160(−4)3 − 2(−4) = 10240 + 8 = 10248 Since 𝑓 ′′ (−4) > 0 then f(x) is convex at 𝑥 = −4 Question #5 Is 𝑓(𝑥) = −9𝑥 5 + 15𝑥 4 − 21𝑥 2 − 8𝑥 + 2 concave or convex at 𝑥 = 3 ? Answer: Step 1 To test for concavity/convexity , require to evaluate f''(x) If 𝑓 ′′ (𝑥) > 0 then is convex If f''(x) then f(x) is concave NKS require to find f''(x) and then value of f''(3) 𝑓 ′ (𝑥) = −45𝑥 4 + 60𝑥 3 − 42𝑥 − 8 𝑓 ′′ (𝑥) = −180𝑥 3 + 180𝑥 2 − 42 𝑓 ′′ (3) = −180(3)3 + 180(3)2 − 42 = −4860 + 1620 − 42 = −3282 < 0 Since f''(3)<0 then f(x) is concave at 𝑥 = 3 Question #6 Is 𝑓(𝑥) = −𝑥 3 − 𝑥 2 + 𝑥 concave or convex at 𝑥 = 4 ? 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 Step 2 Hence: 𝑓(𝑥) = −𝑥 3 − 𝑥 2 + 𝑥 𝑓 ′ (𝑥) = −3𝑥 2 − 2𝑥 + 1 and 𝑓 ′′ (𝑥) = −6𝑥 − 2 hence: 𝑓 ′′ (4) = −6(4) − 2 = −26 Since 𝑓 ′′ (4) < 0 then f(x) is concave at 𝑥 = 4 Question #7 Is 𝑓(𝑥) = −𝑥 5 − 2𝑥 2 − 6𝑥 + 3 concave or convex at 𝑥 = −4 ? Answer: Step 1 Calculate the first and second derivatives 𝑓(𝑥) = −𝑥 5 − 2𝑥 2 − 6𝑥 + 3 𝑓 ′ (𝑥) = −5𝑥 4 − 4𝑥 − 6 𝑓 ′′ (𝑥) = −20𝑥 3 − 4 Therefore, when 𝑥 = −4 𝑓 ′′ (−4) = −20 ⋅ (−4)3 − 4 = 1280 − 4 = 1276 As 𝑓 ′′ (𝑥) > 0 , the functiom is convex at 𝑥 = −4 Question #8 Is 𝑓(𝑥) = −𝑥 5 + 3𝑥 4 − 9𝑥 3 − 2𝑥 2 − 6𝑥 concave or convex at 𝑥 = −3 ? Answer: Method used: Second derivative test Take the second derivative of f(x). 𝑓 ′′ (𝑥) = −20𝑥 3 + 36𝑥 2 − 54𝑥 − 4 Then evaluate f(x) at 𝑥 = −3 Answer: 𝑓’‘(−3) = 1022, hence, concave UPWARD Question #9 Is 𝑓(𝑥) = −𝑥 5 + 3𝑥 4 − 9𝑥 3 − 2𝑥 2 − 6𝑥 concave or convex at 𝑥 = 8 ? Answer: Step 1 When looking for the concavity of a function, it's best to find the second derivative, f’‘(x), of the function f(x) When 𝑓’‘(𝑥) < 0, the f(x) is concave When 𝑓’‘(𝑥) > 0 , the f(x) is convex The first derivative of this function is: 𝑓 ′ (𝑥) = −5𝑥 4 + 12𝑥 3 − 27𝑥 2 − 4𝑥 − 6 Step 2 The second derivative is: 𝑓 ′′ (𝑥) = −20𝑥 3 + 36𝑥 2 − 54𝑥 − 4 Plug in 𝑥 = 8 to get: 𝑓 ′′ (8) = −20(8)3 + 36(8)2 − 54(8) − 4 𝑓 ′′ (8) = −8372 Since 𝑓’‘(8) < 0 , the f(x) is concave at 𝑥 = 8 . Question #10 Is 𝑓(𝑥) = −𝑥𝑒 𝑥 + 𝑥 3 − 𝑥 2 concave or convex at 𝑥 = −2 ? Answer: Step 1 First, you would calculate the first and the second derivatives: 𝑓 ′ (𝑥) = (−1 ⋅ 𝑒 𝑥 + (−𝑥) ⋅ 𝑒 𝑥 ) + 3𝑥 2 − 2𝑥 = (−𝑒 𝑥 − 𝑥𝑒 𝑥 ) + 3𝑥 2 − 2𝑥 𝑓’‘(𝑥) = (−𝑒 𝑥 − 𝑒 𝑥 − 𝑥𝑒 𝑥 ) + 6𝑥 − 2 = −2𝑒 𝑥 − 𝑥𝑒 𝑥 + 6𝑥 − 2 Step 2 Then you would calculate 𝑓’‘(−2) = −2𝑒 −2 − (−2)𝑒 −2 + 6(−2) − 2 = −2e−2 +2e−2 − 12 − 2 −14 < 0 Since 𝑓’‘(−2) < 0, the function is concave Question #11 Is 𝑓(𝑥) = (𝑥 − 3)(𝑥 − 2) − 𝑥 2 + 𝑥 3 concave or convex at 𝑥 = −1 ? Answer: Step 1 To determine if a function is concave up or down at a point, you need to examine the sign of the second derivative f’‘(x) at that point. First, multiply out and simplify f(x): 𝑓(𝑥) = 𝑥 2 − 2𝑥 − 3𝑥 + 6 − 𝑥 2 + 𝑥 3 𝑓(𝑥) = 𝑥 3 − 5𝑥 + 6 Step 2 Find f'(x) and then f’‘(x): 𝑓 ′ (𝑥) = 3𝑥 2 − 5 𝑓’‘(𝑥) = 6𝑥 At 𝑥 = −1, 𝑓’‘(−1) = −6 < 0. Since the second derivative is negative at 𝑥 = −1 , that indicates that the method is concave down. Question #12 Is 𝑓(𝑥) = (𝑥 − 3)(𝑥 − 2) − 𝑥 2 concave or convex at 𝑥 = 3 ? Answer: Explanation: Simplify to see that 𝑓(𝑥) = −5𝑥 + 6. If more is needed, observe that 𝑓 ′ (𝑥) = −5 and 𝑓’‘(𝑥) = 0 , so f is neither concave nor convex. Question #13 Is 𝑓(𝑥) = (𝑥 − 3)(𝑥 + 2)(𝑥 − 4)2 concave or convex at 𝑥 = −1 ? Answer: Step 1 We need (𝑢𝑣𝑤)′ = 𝑢′ 𝑣𝑤 + 𝑣 ′ 𝑢𝑤 + 𝑤 ′ 𝑢𝑣 We must calculate the first and second derivatives 𝑓(𝑥) = (𝑥 − 3)(𝑥 + 2)(𝑥 − 4)2 𝑓 ′ (𝑥) = (𝑥 + 2)(𝑥 − 4)2 + (𝑥 − 3)(𝑥 − 4)2 + 2(𝑥 − 3)(𝑥 + 2)(𝑥 − 4) = (𝑥 − 4){(𝑥 + 2)(𝑥 − 4) + (𝑥 − 3)(𝑥 − 4) + 2(𝑥 − 3)(𝑥 + 2)} = (𝑥 − 4){𝑥 2 − 2𝑥 − 8 + 𝑥 2 − 7𝑥 + 12 + 2𝑥 2 − 2𝑥 − 12} = (𝑥 − 4)(4𝑥 2 − 11𝑥 − 8) 𝑓 ′′ (𝑥) = (4𝑥 2 − 11𝑥 − 8) + (𝑥 − 4)(8𝑥 − 11) = 4𝑥 2 − 11𝑥 − 8 + 9𝑥 2 − 43𝑥 + 44 = 12𝑥 2 − 54𝑥 + 36 Therefore, 𝑓 ′′ (−1) = 12 + 54 + 36 = 102 As, 𝑓 ′′ (−1) > 0 , we conclude that the function is convex Question #14 Is 𝑓(𝑥) = sin 𝑥 concave or convex at 𝑥 = −1 ? Answer: Step 1 We have to know that 𝑑 𝑑𝑥 sin 𝑥 = cos 𝑥 and 𝑑 𝑑𝑥 cos 𝑥 = − sin 𝑥 . Also recall that concavity and convexity are determined through the sign of the second derivative of a function. First finding the second derivative: 𝑓(𝑥) = sin 𝑥 𝑓 ′ (𝑥) = cos 𝑥 𝑓’‘(𝑥) = − sin 𝑥 Step 2 If 𝑓’‘(−1) >0 , then f is convex (commonly called ‘‘concave up’‘) at 𝑥 = −1 . If 𝑓’‘(−1) < 0 , then f is concave (commonly called ‘‘concave down’‘) at 𝑥 = −1 . We see that 𝑓’‘(−1) = − sin(−1) ≈ 08415 Since 𝑓’‘(−1) > 0 , we see that sin 𝑥 is convex (‘‘concave up’‘) at 𝑥 = −1 . Question #15 Is 𝑓(𝑥) = sin 𝑥 concave or convex at 𝑥 = 𝜋 2 ? Answer: Step 1 To even find concavity, you should know what it means by concave up and concave down. if a functions f’‘(x) for a point, like (0.25, 2.25) is positive, like this graph of 𝑥 3 , but in its 2nd derivative form of 9x then its concave-Up Step 2 If a functions 𝑓’‘(𝑥)for a point, like (0.2, -0.54) is negative, like the graph of 3𝑥 3 + 2𝑥 + 4 , which will turn into -27x then its concave-Down but all of these points and graphs are linear, your looking for an point on an non-linear graph (sin(𝑥)). the second derivative of sin(𝑥) is − sin(𝑥) At 𝑥 = 𝜋 , the point is negative, also known as concave-down. since the graph is going negative at the interval [0, 𝜋] . 2 Question #16 Is 𝑓(𝑥) = sin 𝑥 concave or convex at 𝑥 = Answer: 𝜋 5 ? Step 1 1 f(x) is concave at 𝑥0 if 𝑓(𝑥0 ) > (𝑓(𝑥0 − ℎ) + 𝑓(𝑥0 + ℎ)) and 2 1 f(x) is convex at 𝑥0 if 𝑓(𝑥0 ) < (𝑓(𝑥0 − ℎ) + 𝑓(𝑥0 + ℎ)) . 2 where (𝑥0 − ℎ) and (𝑥0 + ℎ) are two values around 𝑥0 . As 𝜋 5 = 36∘ , let us calculate sin x at {𝑥 = 35∘ , 36∘ , 37∘ } sin 3 5∘ = 0.5735764 sin 3 6∘ = 0.5877853 sin 3 7∘ = 0.6018150 Step 2 Now as 1 2 1 (sin⊕ sin 3 7∘ ) = (0.5735764 + 0.6018150) 2 1 = × 1.1753914 = 0.5876957 < sin 3 6∘ 2 𝑓(𝑥) = sin 𝑥 is concave at 𝜋 5 Further if f(x) can be differentiated twice at 𝑥 = 𝑥0 and 𝑓’‘(𝑥0 ) < 0 , it is concave and if 𝑓’‘(𝑥0 ) > 0 , it is convex. As 𝑓(𝑥) = sin 𝑥 , 𝑓 ′ (𝑥) = cos 𝑥 and 𝑓’‘(𝑥) = − sin 𝑥 and 𝑓’‘(36∘ ) = − sin 3 6∘ = −0.5877853 and hence 𝑓(𝑥) = sin 𝑥 is concave at 𝜋 5 Question #17 Is 𝑓(𝑥) = sin 𝑥 concave or convex at 𝑥 = 3𝜋 2 ? 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) = sin 𝑥 𝑓 ′ (𝑥) = cos 𝑥 and 𝑓 ′′ (𝑥) = − sin 𝑥 Step 2 3𝜋 3𝜋 2 2 ⇒ 𝑓’‘ ( ) = − sin ( ) = 1 . 3𝜋 3𝜋 2 2 Since 𝑓’‘ ( ) > 0 then f(x) is convex at 𝑥 = Question #18 Is 𝑓(𝑥) = 1 − 𝑒 3𝑥 concave or convex at 𝑥 = 2 ? Answer: Find the second derivative at 𝑥 = 2 Explanation: If the second derivative is positive , it's concave up . Likewise, if it is negative , it will be concave down . 𝑓 ′ = −3𝑒 3𝑥 𝑓’‘ = −9𝑒 3𝑥 Therefore, 𝑓’‘(2) < 0 , so it is concave down or convex Question #19 Is 𝑓(𝑥) = 1 − 𝑥 − 𝑒 −3𝑥 𝑥 concave or convex at 𝑥 = 4 ? Answer: Step 1 For 𝑓(𝑥) = 1 − 𝑥 − 𝑓 ′ (𝑥) = −1 − 𝑒 −3𝑥 𝑥 , we have −3𝑥𝑒 −3𝑥 −𝑒 −3𝑥 𝑥2 This simplifies (sort of) to 𝑓 ′ (𝑥) = −1 + 𝑒 −3𝑥 3𝑥 + 1 𝑥2 Therefore, 𝑓’‘(𝑥) = 𝑒 −3𝑥 −3𝑥−2 𝑥3 = 3𝑒 −3𝑥 −3𝑥 − 2 3𝑥 + 1 = 𝑒 −3𝑥 ( =3 ) 3 𝑥 𝑥2 −3𝑥 − 2 −9𝑥 − 3 = 𝑒 −3𝑥 ( + ) 𝑥3 𝑥2 = 𝑒 −3𝑥 ( = 𝑒 −3𝑥 ( −3𝑥 − 2 −9𝑥 2 − 3𝑥 + ) 𝑥3 𝑥3 −9𝑥 2 −6𝑥−2 𝑥3 Step 2 Now let 𝑥 = 4 ) 3𝑥+1 𝑥2 𝑓’‘(4) = 𝑒 −12 ( −9(16)2 − 6(4) − 2 ) 43 Observe that the exponential is always positive. The numerator of the fraction is negative for all positive values of x. The denominator is positive for positive values of x. Therefore, 𝑓’‘(4) < 0 Draw your conclusion about concavity. Question #20 Is 𝑓(𝑥) = 1 − 𝑥 − 𝑒𝑥 𝑥2 concave or convex at 𝑥 = −1 ? Answer: Step 1 The second derivative is determinative of concavity. Start by finding the first derivative. 𝑓 ′ (𝑥) = −1 − 𝑒 𝑥 (𝑥 2 )−𝑒 𝑥 (2𝑥) (𝑥 2 )2 𝑓 ′ (𝑥) = 1 − 𝑥(𝑥𝑒 𝑥 − 2) 𝑥4 𝑓 ′ (𝑥) = 1 − 𝑥𝑒 𝑥 − 2 𝑥3 Now differentiate this. 𝑓’‘(𝑥) = 0 − (𝑒 𝑥 + 𝑥𝑒 𝑥 )𝑥 3 − (𝑥𝑒 𝑥 − 2)(3𝑥 2 ) (𝑥 3 )2 𝑓’‘(𝑥) = − 𝑥 3 𝑒 𝑥 + 𝑥 4 𝑒 𝑥 − 3𝑥 3 𝑒 𝑥 + 6𝑥 2 𝑥6 𝑓’‘(𝑥) = − 𝑥 4 𝑒 𝑥 − 2𝑥 3 𝑒 𝑥 + 6𝑥 2 𝑥6 Step 2 We now evaluate the point within the second derivative. 𝑓’‘(−1) = − (−1)4 𝑒 −1 −2(−1)3 𝑒 −1 +6(−1)2 (−1)6 1 −1 −2 +6 𝑒 𝑒 𝑓’‘(−1) = − 1 1 2 + +6 𝑓’‘(−1) = − 𝑒 𝑒 1 This is obviously negative, therefore, the function is convex at 𝑥 = −1 . If the second derivative had had a positive value at 𝑥 = −1 , then we would have dubbed it concave at that point.