If the tangent line to y = f(x) at (3, 2) passes through the point (0, 1), find f(3) and f'(3)?

Solution:

When x = 3 then f(x) = 2 since the point (3,2) lies on the curve.

In other words, the point (3,2) lies on the curve represented by the function y = f(x). So f(3) =2.

f’(x) is the first derivative of the of f(x) and is also the slope of f(x).

At point (3,2) the tangent touches f(x).

At (3,2) the slope of the tangent is the same as the slope of f(x).

The slope of the tangent can be calculated from the two points (3,2) and (0,1) which lie on the tangent. The slope of the tangent is

m = (y₂- y₁)/(x₂ - x₁) = (1 - 2)/(0 - 3) = -1/-3 = 1/3

At (3, 2) the slope of f(x) is f’(3) and is the same as that of the tangent.

f’(3) = 1/3

If the tangent line to y = f(x) at (3, 2) passes through the point (0, 1), find f(3) and f'(3)?

Summary:

A tangent line to y = f(x) at (3, 2) passes through the point (0, 1), find f(3) and f'(3). The values of f(3) and f'(3) are 2 and 1/3 respectively.