The first derivative of a continuous function y =f(x) is y' = (5x2 - 8x) (x- 4)2. Find y" and then use the graphing procedure to sketch the general shape of the graph of f. v"=D View an example | 14 parts remaining x r P The first derivative of a continuous function y =f(x) is y' = (5x2 - 14x) (x 7)2. Find y" and then use the graphing procedure to sketch the general shape of the graph off. (3 Note that y' is a polynomial function and is defined for all real numbers. Since y' is the derivative of a continuous function, the domain of the function is the set of all real numbers. To differentiate y', use the derivative product rule. The derivative product rule states that if u and v are differentiable at x, then the derivative of their product is as follows. d dv du amv): ua +va To find y", differentiate y' using the product rule. Differentiate (x - 7)2. If_i 2_ _ 2 y -dx[(5x 14x)(x 7)] d d = (5X2 '14X) 50 - 7)2 + (X- #3 (5x2 14x) d =2(5x2 - 14x) (x- 7) + (x - 7023 (5x2 - 14x) Differentiate 5x2 - 14x. cl y" =2(5x2 14x) (x- 7) + (x - #5 (5x2 - 14x) =2(5x2 - 14x) (x-7)+ (x- 7)2(1Ox- 14) Factor out 2(x - 7) and simplify the remaining expression. y" =2(5x2 - 14x) (x - 7) +(x- 7)2(10x-14) =2(x - 7) (10x2 -56x + 49) Therefore, the second derivative y" is 2(x - 7) (10x2 - 56x + 49) . To find the critical points, if any, let the first derivative equal 0, and solve for x. (5x2 - 14x) (x- 7)2 = 0 x(5x - 14)(x - 7)2 = 0 X =0 or 5x - 14 = 0 or ( x - 7)2 = 0 14 X = 0 X = or 5 X =7 14 Since y' exists over the domain of y, the critical points are at x = 0, x = -, and x = 7. To determine the behavior at the critical points, use the second derivative test. Suppose f' is continuous on an open interval that contains a critical point c. (1) If f"'(c) 0, then f has a local minimum at x = C. (3) If f"(c) = 0, then the test fails. The function f may have a local maximum, a local minimum, or neither. Determine the behavior of the function at x = 0.y has a local maximum at x = 0. 14 Determine the behavior of the function at x = ?' 14 y has a local minimum at x = ?. Determine the behavior of the function at x = 7. y may have a local maximum, 8 local minimum, or neither at x = 7. Find where the curve is increasing and where it is decreasing. The critical points subdivide the domain of y to create nonoverlapping open intervals on which either positive or negative. Determine the sign of y' over these intervals. 14 14 Interval x 0 at any point in an open interval, then the curve is increasing on that interval. If y' 0 at any point in an open interval, then the curve is concave up on that interval. If y"