Question

1) the equation of a curve is given as f(x)=√2x-1.find the equation of the tangent line to the graph of f(x) at x=5.

2) find the equation of the tangent line to the curve g(x)=x/x+1 at the point (2,2/3).

Answer 1

(1) Given curve is-

$y=\sqrt{2x-1}$

$\Rightarrow y'(x)=\dfrac{2}{2\sqrt{2x-1}}\\[9pt]\Rightarrow y'(x)=\dfrac{1}{\sqrt{2(5)-1}}=\dfrac{1}{\sqrt{9}}=\dfrac{1}{3}$

Also, at $x=5, y=\sqrt{2(5)-1}=\sqrt{10-1}=\sqrt{9}=3$

So Tangent Equation is-

$(y-3)=y'(x)(x-5)\\[9pt] \Rightarrow y-3=\dfrac{x-5}{3}\\[9pt]\Rightarrow 3y-9=x-5\\[9pt]\Rightarrow 3y-x-4=0$

(2) Given curve is-

$g(x)=\dfrac{x}{x+1}$

$g'(x)=\dfrac{(x+1)-x}{(x+1)^2}\\[9pt]\Rightarrow g'(x)_{(2,\frac{2}{3})}=\dfrac{1}{(2+1)^2}=\dfrac{1}{9}$

Tangent Equation is-

$(y-\dfrac{2}{3})=g'(x)(x-2)\\[9pt]\Rightarrow \dfrac{3y-2}{3}=\dfrac{x-2}{9}\\[9pt]\Rightarrow 9y-6=x-2\\[9pt]\Rightarrow x-9y+4=0$