How do you find the equation of the line tangent to #f(x)= (sqrtx+1)#, at (0,1)?

Question

William Abdalla · Accepted Answer

Find the derivative at the point (= the slope of the graph's tangent). Then find the equation with the slope and point that you have.

#=1/2*x^(-1/2)# #=1/2*1/sqrt(x)# #=1/(2sqrt(x))# The slope of the line tangent to f(x) at #(0,1)=# #1/(2sqrt(0))=1/0=>#the line is parallel to the #y#-axis Second: the equation of the tangent is #x=a# where #a# is constant again it passes through (0,1) so equation should be #x=0# i.e. #y# axis

Adam Adkins · Answer

undefined To find the equation of the line tangent to the function f(x) = √x + 1 at the point (0,1), we need to find the slope of the tangent line and the point of tangency.

First, we find the derivative of f(x) with respect to x:
f'(x) = 1/(2√x)

Next, we substitute x = 0 into the derivative to find the slope of the tangent line at (0,1):
f'(0) = 1/(2√0) = undefined

Since the derivative is undefined at x = 0, we cannot directly find the slope. However, we can use the limit definition of the derivative to find the slope:

lim(x→0) [f(x) - f(0)] / (x - 0)

Substituting f(x) = √x + 1 and f(0) = 1, we have:
lim(x→0) [√x + 1 - 1] / x

Simplifying, we get:
lim(x→0) √x / x

Using the limit properties, we can rewrite this as:
lim(x→0) (1/√x) / (1/x)

Simplifying further, we have:
lim(x→0) x / √x

Applying L'Hôpital's rule, we differentiate the numerator and denominator:
lim(x→0) 1 / (1/2√x)

Simplifying, we get:
lim(x→0) 2√x

Substituting x = 0, we find the slope of the tangent line:
f'(0) = 2√0 = 0

Therefore, the slope of the tangent line at (0,1) is 0.

Since the slope is 0, the equation of the tangent line is simply the equation of the horizontal line passing through the point (0,1):

y = 1