Question

6 22 + 3 6) Of all lines tangent to the graph of y = maximum slope. find the tangent lines of minimum slope and

Answer 1

The given function is f(x) = y= 6 22 +3 f'(x) = 12x (x2+3)2 . g(x) = f'(x) = - Let 12x (x2+3)2 When the tangent line has the minimum or maximum slope the first derivative of the function representing the tangent lines (i.e. g(x) ) should be equal to zero. 介 g'(x) =-* 12(x2+3)2 – 48x2(x2+3) (x2+3)4 = + 12(x² + 3)2 – 48x? (x2 + 3) = 0. 36(x2 + 3)(1 - 2) = 0. Assuming that the domain of the function is the set of real numbers, we get, x = = +1. g(-1) It can be seen that -12 (1+3)2 :) = 12 g(1) = and (a 3 4 (1+3)

介 The value of the derivative of the function f(x) is maximum at x = -1 and minimum at x = 1. 6 When X = -1, y = 1+3 3 When x = 1, y= * = 6 1+3 The tangent line having the maximum slope has a slope of and passes through point (-1, 3). 3 The equation of the tangent line having 3 Y- 2 3 the maximum slope is 2+1 3x – 4y + 9 = 0. The tangent line having the minimum slope has a slope of and passes through point (1,7). The equation of the tangent line having Y- the minimum slope is col 3 4 x-1 3x + 4y – 9 = 0.