6 Determine the interval(s) over which the graph of f(x) = x + x 25x+18 is...

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6 Determine the interval(s) over which the graph of f(x) = x + x 25x+18 is concave up or concave down. Concave up on (-5,0) Concave down on Concave up on (-, - 7) U(0, ) ( - /7, 0)uo, c Concave down on Concave up on n(- (-, - /7)U(0, ) (- 5/7, 0) Concave down on man Concave up on on (-0 - -/-) 1 Concave down on From the graph of f(x), determine the graph of f'(x). -6 -5 & -3 -2 -1 y 6.0 5 + 2 O -1 -2 -3 -4 -5 -6 2 3 4 6 X O -6 -5 -4 -3 -2 y 6 5 4 3 2 -2 -3 -4 -5 -6 2 3 4 5 6 X OT O -6 -5 -4 -3 -2 y 6 5 4 3 2 1 -2 -3 -4 -5 -6 2 3 4 5 6 X -6 -5 -4 -3 -2 6 5 4 3 2 O -2 -3 -4 -5 -6 2 3 4 5 9 x This O -6 -5 -4 -3 6 5 4 3 2 5 -2 -3 -4 -5 -6 ~ 3 A 5 6 X This Find the global maximum and minimum points of the function f(x) = 2x 54x + 12 on the interval [0,7]. The global minimum is 12 at x = - 3. The global maximum is 320 at x = 7. O The global minimum is -96 at x =3. The global maximum is 12 at x = - 3. O The global minimum is -96 at x = 3. The global maximum is 320 at x = 7. O The global minimum is 12 at x =0. The global maximum is 320 at x = 7. Determine if the conditions of the mean value theorem are met by the function f(x) = 2x-4x+5 on [0, 3]. If so, find the values of cin (0, 3) guaranteed by the theorem. f(x) is a polynomial and therefore continuous on [0, 3] and differentiable on the interval (0, 3). O The value guaranteed by the mean value theorem is c = 3. O f(x) is a polynomial and therefore continuous on [0, 3] and differentiable on the interval (0, 3). The value guaranteed by the mean value theorem is C = 6 3 f(x) is a polynomial and therefore continuous on [0, 3] and differentiable on the interval (0, 3). The values guaranteed by the mean value theorem are c = - 6 3 6 and c= = f(x) is a polynomial and therefore continuous on [0, 3] and differentiable on the interval (0, 3). The values guaranteed by the mean value theorem are c = - -3 and c = 3. 6 Determine the interval(s) over which the graph of f(x) = x + x 25x+18 is concave up or concave down. Concave up on (-5,0) Concave down on Concave up on (-, - 7) U(0, ) ( - /7, 0)uo, c Concave down on Concave up on n(- (-, - /7)U(0, ) (- 5/7, 0) Concave down on man Concave up on on (-0 - -/-) 1 Concave down on From the graph of f(x), determine the graph of f'(x). -6 -5 & -3 -2 -1 y 6.0 5 + 2 O -1 -2 -3 -4 -5 -6 2 3 4 6 X O -6 -5 -4 -3 -2 y 6 5 4 3 2 -2 -3 -4 -5 -6 2 3 4 5 6 X OT O -6 -5 -4 -3 -2 y 6 5 4 3 2 1 -2 -3 -4 -5 -6 2 3 4 5 6 X -6 -5 -4 -3 -2 6 5 4 3 2 O -2 -3 -4 -5 -6 2 3 4 5 9 x This O -6 -5 -4 -3 6 5 4 3 2 5 -2 -3 -4 -5 -6 ~ 3 A 5 6 X This Find the global maximum and minimum points of the function f(x) = 2x 54x + 12 on the interval [0,7]. The global minimum is 12 at x = - 3. The global maximum is 320 at x = 7. O The global minimum is -96 at x =3. The global maximum is 12 at x = - 3. O The global minimum is -96 at x = 3. The global maximum is 320 at x = 7. O The global minimum is 12 at x =0. The global maximum is 320 at x = 7. Determine if the conditions of the mean value theorem are met by the function f(x) = 2x-4x+5 on [0, 3]. If so, find the values of cin (0, 3) guaranteed by the theorem. f(x) is a polynomial and therefore continuous on [0, 3] and differentiable on the interval (0, 3). O The value guaranteed by the mean value theorem is c = 3. O f(x) is a polynomial and therefore continuous on [0, 3] and differentiable on the interval (0, 3). The value guaranteed by the mean value theorem is C = 6 3 f(x) is a polynomial and therefore continuous on [0, 3] and differentiable on the interval (0, 3). The values guaranteed by the mean value theorem are c = - 6 3 6 and c= = f(x) is a polynomial and therefore continuous on [0, 3] and differentiable on the interval (0, 3). The values guaranteed by the mean value theorem are c = - -3 and c = 3.