How do you find the equation of a line tangent to the function #y=sqrt(2x+1)# at x=4?

Question

Luke Alvarez · Accepted Answer

The tangent has equation #y = 1/3x + 5/3#.

Start by finding the corresponding y-coordinate. #y = sqrt(2(4) + 1) = sqrt(9) = 3# Find the derivative. Let #y= sqrt(u)# and #u = 2x + 1#. Then #dy/(du) = 1/(2sqrt(u))# nad #(du)/dx = 2#. By the chain rule, #dy/dx = dy/(du) * (du)/dx = 1/(2sqrt(u)) * 2 = 1/sqrt(u) = 1/sqrt(2x + 1)#. Find the slope of the tangent by evaluating the point within the derivative. #y= 1/sqrt(2(4) + 1) = 1/sqrt(9) = 1/3# The equation of the tangent is therefore: #y - y_1 = m(x - x_1)# #y - 3 = 1/3(x - 4)# #y -3 = 1/3x - 4/3# #y = 1/3x - 4/3 + 3# #y = 1/3x + 5/3# Hopefully this helps!

Emily Ammerman · Answer

undefined To find the equation of a line tangent to the function y = √(2x + 1) at x = 4, we need to determine the slope of the tangent line and the point of tangency.

First, we find the derivative of the function y = √(2x + 1) with respect to x.

dy/dx = (1/2) * (2x + 1)^(-1/2)

Next, we substitute x = 4 into the derivative to find the slope of the tangent line at x = 4.

dy/dx = (1/2) * (2(4) + 1)^(-1/2) = (1/2) * (9)^(-1/2) = 1/6

So, the slope of the tangent line at x = 4 is 1/6.

To find the point of tangency, we substitute x = 4 into the original function y = √(2x + 1).

y = √(2(4) + 1) = √(9) = 3

Therefore, the point of tangency is (4, 3).

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope of the tangent line, we can substitute the values to find the equation of the tangent line.

y - 3 = (1/6)(x - 4)

Simplifying the equation, we get:

y = (1/6)x + 2/3