QUESIION 9
For certain function f , the first derivative is given as
9.1 Calculate the -coordinates of the stationary points of
9.2 For which values of is concave up
9.3 Determine the values of for which is strictly increasing?
9.4 If it is further given that and , determine the equation ol

Ask by Bartlett Salinas. in South Africa

Feb 11,2025

Question

QUESIION 9
For certain function f , the first derivative is given as
9.1 Calculate the -coordinates of the stationary points of
9.2 For which values of is concave up
9.3 Determine the values of for which is strictly increasing?
9.4 If it is further given that and , determine the equation ol

Ask by Bartlett Salinas. in South Africa

Feb 11,2025

UpStudy AI · Accepted Answer

We are given that f ′(x) = 3x² + 8x – 3. Let's work through each part: 1. For the stationary points, we set f ′(x) = 0. That is, solve: 3x² + 8x – 3 = 0. Using the quadratic formula: x = [–8 ± √(8² – 4·3·(–3))] / (2·3) = [–8 ± √(64 + 36)] / 6 = [–8 ± √100] / 6 = [–8 ± 10] / 6. Thus, the two solutions are: x = (–8 + 10)/6 = 2/6 = 1/3, x = (–8 – 10)/6 = –18/6 = –3. So, the stationary points occur at x = –3 and x = 1/3. 2. The function f is concave up when its second derivative is positive. Differentiate f ′(x) to obtain f ″(x): f ″(x) = d/dx (3x² + 8x – 3) = 6x + 8. Set f ″(x) > 0: 6x + 8 > 0 ⟹ 6x > –8 ⟹ x > –8/6 ⟹ x > –4/3. Thus, f is concave up for all x > –4/3. 3. f is strictly increasing where f ′(x) > 0. Recall f ′(x) = 3x² + 8x – 3, a quadratic with zeros at x = –3 and x = 1/3. Because the coefficient of x² is positive, the quadratic is positive outside the interval between the roots. Thus, f ′(x) > 0 when x < –3 or x > 1/3. 4. We’re also given that f(x) = ax³ + bx² + cx + d, and f(0) = –18. Since f(0) = d, we immediately have: d = –18. Since the derivative f ′(x) is given by: f ′(x) = 3ax² + 2bx + c, and it must equal 3x² + 8x – 3, we equate coefficients: Coefficient of x²: 3a = 3 ⟹ a = 1. Coefficient of x: 2b = 8 ⟹ b = 4. Constant term: c = –3. Thus, we find: a = 1, b = 4, c = –3, d = –18. So, the explicit form of f is: f(x) = x³ + 4x² – 3x – 18. To summarize: 9.1 The stationary points occur at x = –3 and x = 1/3. 9.2 f is concave up for x > –4/3. 9.3 f is strictly increasing where x < –3 or x > 1/3. 9.4 The equation of f is f(x) = x³ + 4x² – 3x – 18. 9.1 The stationary points are at x = –3 and x = 1/3. 9.2 The function f is concave up for all x > –4/3. 9.3 f is strictly increasing when x < –3 or x > 1/3. 9.4 The equation of f is f(x) = x³ + 4x² – 3x – 18.

QUESIION 9
For certain function f , the first derivative is given as
9.1 Calculate the -coordinates of the stationary points of
9.2 For which values of is concave up
9.3 Determine the values of for which is strictly increasing?
9.4 If it is further given that and , determine the equation ol

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QUESIION 9 For certain function f , the first derivative is given as 9.1 Calculate the -coordinates of the stationary points of 9.2 For which values of is concave up 9.3 Determine the values of for which is strictly increasing? 9.4 If it is further given that and , determine the equation ol

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QUESIION 9
For certain function f , the first derivative is given as
9.1 Calculate the -coordinates of the stationary points of
9.2 For which values of is concave up
9.3 Determine the values of for which is strictly increasing?
9.4 If it is further given that and , determine the equation ol