Question

Find the tangent line approximation for 10+x−−−−−√ near x=0. Do not approximate any of the values in your formula when entering your answer below.

Answer 1

Answer:

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Step-by-step explanation:

We are asked to find the tangent line approximation for $f(x)=\sqrt{10+x}$ near $x=0$.

We will use linear approximation formula for a tangent line $L(x)$ of a function $f(x)$ at $x=a$ to solve our given problem.

$L(x)=f(a)+f'(a)(x-a)$

Let us find value of function at $x=0$ as:

$f(0)=\sqrt{10+x}=\sqrt{10+0}=\sqrt{10}$

Now, we will find derivative of given function as:

$f(x)=\sqrt{10+x}=(10+x)^{\frac{1}{2}}$

$f'(x)=\frac{d}{dx}((10+x)^{\frac{1}{2}})\cdot \frac{d}{dx}(10+x)$

$f'(x)=\frac{1}{2}(10+x)^{-\frac{1}{2}}\cdot 1$

$f'(x)=\frac{1}{2\sqrt{10+x}}$

Let us find derivative at $x=0$

$f'(0)=\frac{1}{2\sqrt{10+0}}=\frac{1}{2\sqrt{10}}$

Upon substituting our given values in linear approximation formula, we will get:

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}(x-0)$  

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}x-0$

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Therefore, our required tangent line for approximation would be $L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$.