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

Question

Evelyn Agard · Accepted Answer

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

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#

Paisley Johnson · Answer

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

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#

Elijah Abram · Answer

undefined To find the equation of the line tangent to the graph of f(x) = √(x-1) at the point (5,2), we need to determine the slope of the tangent line at that point.

First, we find the derivative of f(x) with respect to x.

f'(x) = 1 / (2√(x-1))

Next, we substitute x = 5 into the derivative to find the slope at the point (5,2).

f'(5) = 1 / (2√(5-1)) = 1 / (2√4) = 1/4

So, the slope of the tangent line at (5,2) is 1/4.

Using the point-slope form of a linear equation, we can write the equation of the tangent line as:

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

Simplifying this equation, we get:

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

Finally, rearranging the equation, we have:

y = (1/4)x - 5/4 + 2

y = (1/4)x - 5/4 + 8/4

y = (1/4)x + 3/4

Therefore, the equation of the line tangent to the graph of f(x) = √(x-1) at the point (5,2) is y = (1/4)x + 3/4.