Question

How do you find the equation of the line tangent to the graph of `f(x)=sqrt( x-1)` at the point (5,2)?

Answer 1

First, find the slope. (We already have point (5, 2).)

Explanation:

I will assume that you have not yet been taught the rules for finding derivatives (the 'shortcuts'). So, we will use a definition.

The slope of the line tangent to the graph of the function `f` at the point `(a, f(a))` can be defined in several ways (or using several notations). Two of the more common are:

`lim_(xrarra) (f(x)-f(a))/(x-a)` `" "` OR `" "` `lim_(hrarr0) (f(a+h)-f(a))/h`

(Each author,teacher,presenter needs to choose one definition as the 'official' definition. Many will immediately mention other possibilities as 'equivalents'.)

For this question we have `f(x) = sqrt(x-1)` and `a=5`

We'll find:

`lim_(hrarr0) (f(5+h)-f(5))/h`
(Note that substitution gets us the indeterminate form `0/0`. We have some work to do.)

`lim_(hrarr0) (f(5+h)-f(5))/h = lim_(hrarr0) (sqrt((5+h)-1)-sqrt((5)-1))/h`

`= lim_(hrarr0) (sqrt(4+h)-2)/h`

`= lim_(hrarr0) ((sqrt(4+h)-2))/h * ((sqrt(4+h)+2))/((sqrt(4+h)+2))`

`= lim_(hrarr0) (4+h-4)/(h(sqrt(4+h)+2))`

`= lim_(hrarr0) h/(h(sqrt(4+h)+2))`

The expression whose limit we want is equal to `1/(sqrt(4+h)+2)` for all `h` other than `h=0`. The limit doesn't care what happens when `h` is equal to `0`, it wants to know what happens when `h` is close to `0`, so we get:

`lim_(hrarr0) h/(h(sqrt(4+h)+2)) = lim_(hrarr0) 1/(sqrt(4+h)+2) = 1/(sqrt4 =2) = 1/4`

The slope of the tangent we were asked about is `1/4`.

So the tangent line has slope `1/4` and includes the point `(5,2)`.

The equation, in slope-intercept form, of that line is:

`y = 1/4 x +3/4`

If you already know the power rule and the chain rule,

find `f'(x) = 1/(2sqrt(x-1))`,

so the slope of the tangent line at `x=5` is `f'(5) = 1/4`

Answer 2

`y = 1/4x +3/4`

Explanation:

The slope of the tangent to the graph of `f(x) = sqrt(x-1)` is given by the first derivative of the function, `f^(')(x)`.

`color(blue)("slope" = m = f^(')(x))`

You can aclculate it by using the chain rule for `u^(1/2)`, with `u = (x-1)`.

`d/dx(u^(1/2)) = d/(du)u^(1/2) * d/dx(u)`

`d/dx(u^(1/2)) = 1/2 u^(-1/2) * d/dx(x-1)`

`d/dx((x-1)^(1/2)) = 1/2 * 1/sqrt(x-1) * 1`

`d/dx(sqrt(x-1)) = 1/(2sqrt(x-1))`

For your point `(5,2)`, the slope will be equal to

`m = 1/(2 * sqrt(5-1)) = 1/(2 * 2) = 1/4`

The point-slope form for a general line given a point `(x_1,y_1)` is

`color(blue)(y-y_1 = m * (x - x_1))`

For your point, you have

`y - 2 = 1/4 * (x - 5)`

Alternatively, you can rewrite this in slope-intercept form

`y = 1/4x - 5/4 + 8/4`

`y = 1/4x +3/4`