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

Dec 25, 2015

$y = 1.307 x - 2.614$

#### Explanation:

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

$y - {y}_{1} = m \left(x - {x}_{1}\right)$ ,

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 $f \left(x\right)$ to get a corresponding ${y}_{1}$ value.

$f \left(2\right) = \left(2 - 3\right) \left(2 - 2\right) \left(\ln \left(2\right) - 2\right) = 0 = {y}_{1}$

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.

$f \left(x\right) = \left(x - 3\right) \left(x - 2\right) \left(\ln x - x\right)$

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:

$\frac{d}{\mathrm{dx}} \left[f \left(x\right) g \left(x\right) h \left(x\right)\right] =$
$f ' \left(x\right) g \left(x\right) h \left(x\right) + f \left(x\right) g ' \left(x\right) h \left(x\right) + f \left(x\right) g \left(x\right) h ' \left(x\right)$

$f ' \left(x\right) = \left(1\right) \left(x - 2\right) \left(\ln x - x\right) + \left(x - 3\right) \left(1\right) \left(\ln x - x\right) + \left(x - 3\right) \left(x - 2\right) \left({x}^{-} 1 - 1\right)$

$m = f ' \left(2\right) = 0 + \left(2 - 3\right) \left(1\right) \left(\ln 2 - 2\right) + 0$

$m = - \left(\ln 2 - 2\right) = 2 - \ln 2 \approx 1.307$

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

$y - 0 = \left(1.307\right) \left(x - 2\right)$
$y = 1.307 x - \left(2\right) \left(1.307\right)$
$y = 1.307 x - 2.614$