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

Jul 30, 2018

$y = \left(- \frac{2}{3} + 2 \left(I n 3 - 3\right)\right) x - 5 I n 3 + 17$

#### Explanation:

$f \left(x\right) = \left(x - 2\right) \left(x - 2\right) \left(I n x - x\right)$
$f \left(x\right) = {\left(x - 2\right)}^{2} \left(I n x - x\right)$
$f ' \left(x\right) = {\left(x - 2\right)}^{2} \left(\frac{1}{x} - 1\right) + \left(I n x - x\right) \times 2 \left(x - 2\right)$
$f ' \left(x\right) = {\left(x - 2\right)}^{2} \left(\frac{1}{x} - 1\right) + 2 \left(x - 2\right) \left(I n x - x\right)$

At $x = 3$,
$f ' \left(3\right) = {\left(3 - 2\right)}^{2} \left(\frac{1}{3} - 1\right) + 2 \left(3 - 2\right) \left(I n 3 - 3\right)$
$f ' \left(3\right) = - \frac{2}{3} + 2 \left(I n 3 - 3\right)$

Equation of the tangent at $\left(3 , I n 3 - 3\right)$

$y - \left(I n 3 - 3\right) = \left(- \frac{2}{3} + 2 \left(I n 3 - 3\right)\right) \left(x - 3\right)$

$y - I n 3 + 3 = \left(- \frac{2}{3} + 2 \left(I n 3 - 3\right)\right) x + 2 - 6 \left(I n 3 - 3\right)$

$y - I n 3 + 3 = \left(- \frac{2}{3} + 2 \left(I n 3 - 3\right)\right) + 2 - 6 I n 3 + 18$

$y - I n 3 = \left(- \frac{2}{3} + 2 \left(I n 3 - 3\right)\right) x - 6 I n 3 + 17$

$y = \left(- \frac{2}{3} + 2 \left(I n 3 - 3\right)\right) x - 5 I n 3 + 17$

Jul 30, 2018

Equation of tangent is $y = - 4.47 x + 11.5$

#### Explanation:

f(x)= (x-2)(x-2)(ln x- x) ; x=3 or

$f \left(x\right) = {\left(x - 2\right)}^{2} \left(\ln x - x\right)$

$f \left(3\right) = {\left(3 - 2\right)}^{2} \left(\ln 3 - 3\right) \approx - 1.90$ The point at which ,tangent

to be drawn is $\left(3 , - 1.9\right)$

$f \left(x\right) = {\left(x - 2\right)}^{2} \left(\ln x - x\right)$

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

$f ' \left(3\right) = {\left(3 - 2\right)}^{2} \left(\frac{1}{3} - 1\right) + 2 \left(3 - 2\right) \left(\ln 3 - 3\right) \approx - 4.47$

Slope of curve at $\left(3 , - 1.9\right)$ is $m = - 4.47$

Equation of tangent is $y - {y}_{1} = m \left(x - {x}_{1}\right)$ , therefore

equation of tangent is $y - \left(- 1.9\right) = - 4.47 \left(x - 3\right)$ or

$y + 1.9 = - 4.47 x + 13.4 \mathmr{and} y = - 4.47 x + 11.5$ [Ans]