How do you use the limit definition to find the slope of the tangent line to the graph f(x)= x(sqrt(x)-1) at x=4?

1 Answer
Nov 8, 2016

graph{(x(sqrt(x)-1)-y)(2x-y-4)=0 [-1.99, 13.814, -0.53, 7.37]}

2x-y-4=0

Explanation:

The tangent is given by

y-f(x_0)=lim_(h->0)(f(x_0+h)-f(x_0))/h*(x-x_0)

Then

x_0=4\ \ \ \ ; \ \ \ f(x_0)=4

lim_(h->0)(f(x_0+h)-f(x_0))/h=lim_(h->0)((4+h)(sqrt(4+h)-1)-4)/h=
=lim_(h->0)[sqrt(4+h)-1+4(sqrt(4+h)-2)/h]=
=2-1+4lim_(h->0)(4+h-4)/(h*(sqrt(4+h)+2))=2-1+4*1/4=2

So the tangent is

y-4=2(x-4)