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

Question

Scarlett Allison · Accepted Answer

I found: #y=1/3x+2/3#

First you need to find the slope #m# of the tangent line. This is found deriving your function and calculating it at #x=1#:

#f'(x)=(2(x+2)-(2x+1))/(x+2)^2=(2x+4-2x-1)/(x+2)^2=3/(x+2)^2#

at #x=1#
#f'(3)=3/(1+2)^2=1/3=m#

Now we need the #y# value of the tangence point as well; we find it setting #x=1# into the original function:
#f(1)=(2+1)/(1+2)=3/3=1#

The equation of a line passing through #x_0=1 and y_0=1# with slope #m=1/3# is given as:
#y-y_0=m(x-x_0)#
#y-1=1/3(x-1)#
#y=1/3x-1/3+1#
#y=1/3x+2/3#

Graphically:

Evelyn Agard · Answer

undefined The equation of the tangent line at x=1 for the function f(x)=(2x+1)/(x+2) is y=3x-1.

Ezekiel Alexander · Answer

undefined To find the equation of the tangent line at $ x = 1 $ for the function $ f(x) = \frac{2x + 1}{x + 2} $, we need to find the slope of the tangent line at $ x = 1 $ and then use the point-slope form of a line.

1. Find the derivative of the function:
$$ f'(x) = \frac{d}{dx}\left(\frac{2x + 1}{x + 2}\right) $$
Using the quotient rule:
$$ f'(x) = \frac{(2)(x + 2) - (2x + 1)(1)}{(x + 2)^2} $$
$$ f'(x) = \frac{2x + 4 - 2x - 1}{(x + 2)^2} $$
$$ f'(x) = \frac{3}{(x + 2)^2} $$

2. Evaluate the derivative at $ x = 1 $ to find the slope of the tangent line:
$$ f'(1) = \frac{3}{(1 + 2)^2} = \frac{3}{9} = \frac{1}{3} $$

3. Find the value of $ f(1) $:
$$ f(1) = \frac{2(1) + 1}{1 + 2} = \frac{3}{3} = 1 $$

4. Now, we have the slope ($ m = \frac{1}{3} $) and a point ($ x = 1, y = 1 $) on the tangent line. Use the point-slope form of a line to find the equation of the tangent line:
$$ y - y_1 = m(x - x_1) $$
$$ y - 1 = \frac{1}{3}(x - 1) $$
$$ y - 1 = \frac{1}{3}x - \frac{1}{3} $$
$$ y = \frac{1}{3}x + \frac{2}{3} $$

So, the equation of the tangent line at $ x = 1 $ is $ y = \frac{1}{3}x + \frac{2}{3} $.