What is the equation of the tangent line of #f(x)=(x-3)(x-2)(lnx-x)# at #x=2#?
1 Answer
#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
We know
#f(2) = (2-3)(2-2)(ln(2) - 2) = 0 = y_1#
All that remains is to find the slope,
#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
#y-0 = (1.307)(x-2)#
#y=1.307x - (2)(1.307)#
#y=1.307x - 2.614#