If the tangent line to #y = f(x)# at #(4,3)# passes through the point #(0,2)#, Find #f(4)# and #f'(4)#? An explanation would also be very helpful.

Question

Elijah Abram · Accepted Answer

#f(4) = 3#

#f'(4) = 1/4# The question gives you #f(4)# already, because the point #(4,3)# is given. When #x# is #4#, #[y = f(x) = ]f(4)# is #3#. We can find #f'(4)# by finding the gradient at the point #f(4)#, which we can do because we know the tangent touches both #(4,3)# and #(0,2)#. The gradient of a line is given by rise over run, or the change in #y# divided by the change in #x#, or, mathematically, #m = (y_2-y_1)/(x_2-x_1)# We know two points on the graph in the question, so effectively we know the two values we need for #y# and #x# each. Say that #(0,2) -> x_1 = 0, y_1 = 2# #(4,3) -> x_2 = 4, y_2 = 3# so #m = (3-2)/(4-0) = 1/4# which is the gradient.

Aurora Alvarado · Answer

undefined To find f(4) and f'(4), we can use the information given about the tangent line.

First, let's find the slope of the tangent line. The slope of a line passing through two points (x₁, y₁) and (x₂, y₂) is given by (y₂ - y₁) / (x₂ - x₁).

Using the points (4, 3) and (0, 2), the slope of the tangent line is (2 - 3) / (0 - 4) = -1/4.

Since the tangent line is also the derivative of the function f(x) at x = 4, we have f'(4) = -1/4.

Now, let's find f(4) using the point-slope form of a line. The equation of a line passing through the point (x₁, y₁) with slope m is given by y - y₁ = m(x - x₁).

Using the point (4, 3) and the slope -1/4, we have y - 3 = (-1/4)(x - 4).

Simplifying the equation, we get y - 3 = (-1/4)x + 1.

Rearranging the equation, we have y = (-1/4)x + 4.

Therefore, f(4) = 4.

In summary, f(4) = 4 and f'(4) = -1/4.