Let g(x) = f(x) + f(8-x) and f''(x) > 0 for x (0,8) . If g is decreasing in the interval (0, ) and increasing in the interval ( , 8) , then the area of the triangle formed by the tangent to the curve y = x^2 - x + 5 at x = , the normal to the same curve at x = , and the y-axis is

Step-by-step solution

Given g(x) = f(x) + f(8-x) . Differentiating with respect to x , we get: g'(x) = f'(x) - f'(8-x) Since f''(x) > 0 , f'(x) is a strictly increasing function. For g'(x) = 0 , we must have f'(x) = f'(8-x) x = 8-x x = 4 . For x For x > 4 , x > 8-x f'(x) > f'(8-x) g'(x) > 0 . Thus, g(x) is decreasing on (0, 4) and increasing on (4, 8) . Comparing with the given intervals, we get = 4 . The equation of the curve is y = x^2 - 4x + 5 . At x = 4 , y = (4)^2 - 4(4) + 5 = 5 . The point of contact is (4, 5) . Differentiating th