What is the equation of the tangent line of f(x)=xsqrt(x-1) at x=4?

1 Answer
Jun 24, 2018

y=5/sqrt3x-20/sqrt3

Explanation:

f'(x)=x*1/(2sqrt(x-1))+sqrt(x-1)

simplify

f'(x)=x/(2sqrt(x-1))+sqrt(x-1)

f'(x)=x/(2sqrt(x-1))+sqrt(x-1)*(2sqrt(x-1))/(2sqrt(x-1))

f'(x)=x/(2sqrt(x-1))+(2(x-1))/(2sqrt(x-1))

f'(x)=(x+2(x-1))/(2sqrt(x-1))

expand the brackets

f'(x)=(x+2x-2)/(2sqrt(x-1))

f'(x)=(3x-2)/(2sqrt(x-1))

plug into point-slope formula

y-y_1=f'(x_1)(x-x_1)

remembering that the derivative finds the slope of a function so m=f'(x_1)

to find y_1

f(4)=(4)sqrt((4)-1)

y_1=4sqrt(3)

to find f'(x_1)

f'(4)=(3(4)-2)/(2sqrt((4)-1)

f'(4)=(12-2)/(2sqrt(3))

f'(4)=10/(2sqrt(3))

f'(4)=5/sqrt(3)

plug in

y-4sqrt(3)=5/sqrt(3)(x-4)

add 4sqrt(3) to each side

ycancel(-4sqrt(3)+4sqrt(3))=5/sqrt(3)(x-4)+4sqrt3

expand brackets

y=5/sqrt(3)x-5/sqrt3(4)

y=5/sqrt3x-20/sqrt3