For #f(x)=(2x+1)^2/(x-1/2) # what is the equation of the tangent line at #x=2#?

Question

1564 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-05T23:32:12

#y=4/5x+35.9#

You want to use the formula for the tangent line at x=2, which is:

#y=f'(2)(x-2)+f(2)#

First let's find #f(2)# because that is easier:

#f(2)=(2(2)+1)^2/(2-1/2)=25/(3/2)=37.5#

Now #f'(2)# using the quotient rule:

#(u/v)'=(u'v-uv')/v^2# where #u=(2x+1)^2#and #v=x-1/2#

So #u'=2(2x+1)(2)# and #v'=1#

#f'=(2(2x+1)(2)(x-1/2)-(2x+1)^2(1))/(x-1/2)^2#

So plugging in 2 for x:

#f'(2)=(4(5)(1.5)-25)/(1.5)^2=5/6.25=4/5#

Finally our equation will be:

#y=(4/5)(x-2)+37.5->y=4/5x+35.9#