For $f (x) = \frac{{(2 x + 1)}^{2}}{x - \frac{1}{2}}$ $f(x)=(2x+1)^2/(x-1/2)$ what is the equation of the tangent line at $x = 2$ $x=2$ ?

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Answer 1 · 2018-06-05T23:32:12

$y = \frac{4}{5} x + 35.9$ $y=4/5x+35.9$

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:

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

For f(x)=(2x+1)^2/(x-1/2) f(x)=(2x+1)2x−12f(x)=(2x+1)^2/(x-1/2) what is the equation of the tangent line at x=2x=2x=2?