Question

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

Answer 1

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

Explanation:

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!