# Find the tangent line approximation T to the graph f at the given point: f(x) = \sqrt x; (2,...

## Question:

Find the tangent line approximation {eq}T {/eq} to the graph {eq}f {/eq} at the given point:

{eq}f(x) = \sqrt x; (2, \sqrt 2) {/eq}

## Tangent Line Approximation:

The tangent line approximation of a function {eq}y=f(x) {/eq} at the point {eq}(x_0,y_0) {/eq} is given by the equation of the staright line {eq}y-y_0=f'(x_0)(x-x_0). {/eq}

Note that, here {eq}f'(x_0) {/eq} being the slope of the tangent line to the curve.

Given {eq}f(x) = \sqrt x {/eq} and the point {eq}(2, \sqrt 2). {/eq}

The tangent line approximation to the curve at {eq}(2, \sqrt 2) {/eq} is {eq}\displaystyle y-\sqrt{2}=f'(2)(x-2) \quad ---(i). {/eq}

Now,

{eq}\begin{align*} f'(x) &=\frac{d}{dx}\sqrt x\\ \\ &=\frac{1}{2\sqrt{x}}. \end{align*} {/eq}

Then,

{eq}f'(2)=\displaystyle \frac{1}{2\sqrt{2}}. {/eq}

Substituting the value of the derivative in {eq}(i), {/eq} we obtain:

{eq}\displaystyle y-\sqrt{2}= \frac{1}{2\sqrt{2}}(x-2) {/eq}

{eq}\Rightarrow \displaystyle y-\sqrt{2}=\frac{x}{2\sqrt{2}}-\frac{1}{ \sqrt{2}} {/eq}

{eq}\Rightarrow \displaystyle y=\frac{x}{2\sqrt{2}}-\frac{1}{ \sqrt{2}}+\sqrt{2} {/eq}

{eq}\Rightarrow {\color{Blue} { \displaystyle y=\frac{1}{2\sqrt{2}} \ x+\frac{1}{ \sqrt{2}}.}} {/eq} 