Question

How do you find the equation of a line tangent to the function `y=xsqrtx` at (4,8)?

Answer 1

Write the function as:

`y = x^(3/2)`

Verify that the point, `(4,8)`, is on the curve:

`8= 4^(3/2)`

`8=8 larr` verified

Compute the first derivative of the function:

`dy/dx = 3/2x^(1/2)`

The slope, m, of the tangent line is the first derivative evaluated at the x coordinate of the point, `x =4`:

`m = 3/2 4^(1/2)`

`m = 3`

Use the point-slope form of the equation of a line for the tangent line:

`y = m(x-x_0)+y_0`

Substitute `m = 3`, `x_0 = 4` and `y_0 = 8`:

`y = 3(x-4)+8`

Simplify:

`y = 3x-4`

Answer 2

`y=3x-4`

Explanation:

`f(x) = xsqrtx`
[NB: I have used `f(x)` in place of the `y` in the question to avoid confusion with the eauation of the tangent.]

`f(x) =x^1xx^(1/2)=x^(3/2)`

Apply power rule.

`f'(x) = 3/2x^(1/2) = (3sqrtx)/2`

`:.` the slope of `f(x)` at `x=4` is `(3*sqrt4)/2 =+-3`

Since `f(x)` is defined for `x>=0 -> f'(4) =+3`

Let `y=mx +c` be the tangent to `f(x)` at `x=4`

Since `y` at `x=4` will have the same slope as `f(x) -> m=3`

`f(4) = 4 xx sqrt4 =8`

Thus, `8 = 3xx4+c -> c=-4`

Hence, the equation of our tangent line is: `y=3x-4`