Find an equation of the line perpendicular to the graph of F(x)=f(x)+3x2−g(x) at x=1 .

Question

Assume the functions f and g are differentiable functions, with f(1)=6 , f′(1)=−2 , g(1)=−1 and g′(1)=3 .

Answer 1

To find the y-coord of the pt. the line passes thru: F(1) = f(1) + 3·1² - g(1) = 6 + 3 - (-1) = 10 so (1 , 10)

To find the slope of the normal (perpendicular) line: F^'(x) = f^'(x) + 6x - g^'(x) ; F^'(1) = - 2 + 6 - 3 = 1

The slope of the tangent is 1 so the slope of the normal is the opposite reciprocal, so - 1.

Eqn of normal line to graph of F(x) at x = 1: y - 10 = -1(x - 1)