Question

What is the equation of the tangent line of `f(x)=(x-3)(x-2)(lnx-x)` at `x=2`?

Answer 1

`y=1.307x - 2.614`

Explanation:

To find the equation of the tangent line of this function, we must consider the point-slope formula for a line:

`y-y_1 = m(x-x_1)` ,

where `x_1` and `y_1` are known points in the equation, and `m` is the slope at that point.

We know `x_1 = 2`, so we just need to plug that value into `f(x)` to get a corresponding `y_1` value.

`f(2) = (2-3)(2-2)(ln(2) - 2) = 0 = y_1`

All that remains is to find the slope, `m`. Since the derivative of a function represents the slope of that function, we must find the derivative and evaluate it at `x_1` to find the slope at that point.

`f(x) = (x-3)(x-2)(lnx - x)`

To derive this function, we can multiply it out into separate terms and derive them individually with the product rule for 2 terms, but it would be faster to use the product rule for 3 terms:

`d/dx[f(x)g(x)h(x)] =`
`f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)`

`f'(x) =(1)(x-2)(lnx - x) + (x-3)(1)(lnx-x) + (x-3)(x-2)(x^-1 - 1)`

`m = f'(2) = 0 + (2-3)(1)(ln2 - 2) + 0`

`m = -(ln2 - 2) = 2-ln2 ~~ 1.307`

Plugging into our point-slope formula, we arrive at the line approximately tangent to our function at `x=2`:

`y-0 = (1.307)(x-2)`
`y=1.307x - (2)(1.307)`
`y=1.307x - 2.614`