What are the solutions of the equation (x−13)2=400?
1 answer:
You might be interested in
Let . The tangent plane to the surface at (0, 0, 8) is
The gradient is
so the tangent plane's equation is
The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by , then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation
or ,
, and
.
(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)
To solve this we are going to use the average rate of change formula: 
where
is the average rate of change of the function
is the position function evaluated at 
is the position function evaluated at 
is the first point in the interval
is the second point in the interval
We can infer for our problem that the first point is 3 and the second point is 6, so
and 
. Lets replace those values in our formula:
We can conclude that the average rate of change of the function f(x) = 1 x - x2 - 2 for <span>3 ≤ x ≤ 6 is -8</span>
where
We can infer for our problem that the first point is 3 and the second point is 6, so
We can conclude that the average rate of change of the function f(x) = 1 x - x2 - 2 for <span>3 ≤ x ≤ 6 is -8</span>