**Unformatted text preview: **(b) Write an equation for each horizontal tangent line in point-slope form. SOLUTION: (a) Horizontal tangent lines imply slope zero of the function. Take f and set it to zero. Now, f ( x ) = d dx e ln(2) x 2 = e ln(2) x 2 ln(2)(2 x ) = 2 x 2 ln(2)(2 x ) by the chain rule and diﬀerentiation of exponential functions, and 2 x 2 ln(2)(2 x ) = 0 when (2 x ) = 0. Thus, x = 0 is the only value of x so that the tangent line is horizontal. (2 point) (b) Writing this tangent line in point slope form requires a point, ( x ,y ) and a slope m , and the general equation y-y = m ( x-x ). In our problem x = 0 ,y = 2 2 = 2 = 1, and m = 0 because its a horizontal line. Thus, y-1 = 0( x-1), which implies y = 1. (2 points)...

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- Fall '09
- Calculus, Derivative, dt