Question

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?

Answer 1

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)`