Question

find the equation of tangent line to the curve y=(2x+1) /(3x-1) at the point where x=1??

Answer 1

Answer:

x=1

y= (2x+1) /(3x-1)

=2(1) +1/3(1) -1

=2+1/3-1

=3/2

Answer 2

Equation of the curve is,

$\longrightarrow y=\dfrac{2x+1}{3x-1}$

The y coordinate of the point on the curve where $x=1$ is,

$\longrightarrow y=\dfrac{2+1}{3-1}$

$\longrightarrow y=\dfrac{3}{2}$

Slope of tangent to the curve,

$\longrightarrow m=\dfrac{dy}{dx}$

$\longrightarrow m=\dfrac{d}{dx}\left(\dfrac{2x+1}{3x-1}\right)$

$\longrightarrow m=\dfrac{2(3x-1)-3(2x+1)}{(3x-1)^2}$

$\longrightarrow m=-\dfrac{5}{(3x-1)^2}$

At $x=1,$

$\longrightarrow m=-\dfrac{5}{(3-1)^2}$

$\longrightarrow m=-\dfrac{5}{4}$

Therefore, equation of tangent on the curve at the point $\left(1,\ \dfrac{3}{2}\right)$ is, by point - slope form,

$\longrightarrow y-\dfrac{3}{2}=-\dfrac{5}{4}\left(x-1\right)$

$\longrightarrow y-\dfrac{3}{2}=-\dfrac{5x}{4}+\dfrac{5}{4}$

Multiplying by 4,

$\longrightarrow 4y-6=-5x+5$

$\longrightarrow\underline{\underline{5x+4y-11=0}}$