What is the equation of the line tangent to # f(x)=sqrt(5+x) # at # x=-1 #?

Question

What is the equation of the line tangent to #  f(x)=sqrt(5+x) # at #  x=-1 #?

Isaiah Allen · Accepted Answer

Equation of the line is
#x-4y=-9" "#the tangent line

From the given #f(x)=sqrt(5+x) # at #x=-1# We need to solve for the slope at the given point. Obtain the first derivative #f' (x)#, then find slope #m=f' (-1)# The first derivative #d/dx(f(x))=f' (x)=d/dx(sqrt(5+x))# #f' (x)=1/(2sqrt(5+x))*d/dx(5+x)# #f' (x)=1/(2sqrt(5+x))*(d/dx(5)+d/dx(x))# #f' (x)=1/(2sqrt(5+x))*(0+1)# #f' (x)=1/(2sqrt(5+x))# Determine now the slope #m=f' (-1)# #m=f' (-1)=1/(2sqrt(5+(-1)))=1/(2sqrt4)=1/4# We now have the slope. We need to find the ordinate #y=f(-1)# using #x=-1# #f(x)=sqrt(5+x) # #f(-1)=sqrt(5+(-1)) # #f(-1)=sqrt(4) # #f(-1)=2# the required point #(-1, 2)# Now that we have the equation for the tangent line in Point-Slope Form, we can solve it. #y-y_1=m(x-x_1)# #y-2=1/4(x--1)# #4(y-2)=x+1# #4y-8=x+1# #x-4y=-9" "#the tangent line ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Kindly see the graphs of #f(x)=sqrt(5+x)# and the tangent line #x-4y=-9" "# at #(-1, 2)# graph{((y-sqrt(5+x))(x-4y+9))=0[-7,5,-3,3]} May God bless you all. I hope this explanation helps.

Cameron Alexander · Answer

undefined The equation of the line tangent to f(x) = √(5+x) at x = -1 is y = (1/4)x + (3/4).