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

Answer 1

William Abdalla

#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#

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Answer 2

Ezra Adams

To find the equation of the tangent line at x=2, we need to find the derivative of the function f(x) and evaluate it at x=2. The derivative of f(x) is given by f'(x) = (4x+2)/(x-1/2)^2. Evaluating f'(x) at x=2, we get f'(2) = (4(2)+2)/(2-1/2)^2 = 14. Therefore, the equation of the tangent line at x=2 is y = 14(x-2) + f(2), where f(2) is the value of f(x) at x=2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.