Linearization of sqrt(x+1) +sin x at x=0

[karush]

Find the linearization of $f(x) = \sqrt{x+1} +\sin{ x} $ at $x=0$

$L(0)=\frac{3x} {2} + 1$

how is it related to the individual linearization of
$\sqrt{x+1}$ and $\sin{x} $ at $x=0$ ?

$L_{\sqrt{x+1}} (0 ) = \frac{x} {2}+1 $

 

[MarkFL]

Suppose we are given:

\(\displaystyle f(x)=g(x)+h(x)\)

The linearization of $g(x)$ at $x=a$ is found by observing the point $(a,g(a))$ is on the line, and the slope is $g'(a)$, thus the linearization of $g$ at $x=a$ is:

\(\displaystyle y_g=g'(a)(x-a)+g(a)\)

And thus, similarly we have:

\(\displaystyle y_h=h'(a)(x-a)+h(a)\)

And also:

\(\displaystyle y_f=f'(a)(x-a)+f(a)\)

But, we know:

\(\displaystyle f(a)=g(a)+h(a)\) and \(\displaystyle f'(a)=g'(a)+h'(a)\)

And so we may state:

\(\displaystyle y_f=(g'(a)+h'(a))(x-a)+g(a)+h(a)=(g'(a)(x-a)+g(a))+(h'(a)(x-a)+h(a))=y_g+y_h\)

So, we see that the linearization of $f$ is the sum of the linearizations of $g$ and $h$.

 

[karush]

So
$1+\frac{1}{2}=\frac{3}{2}$

Since

$L_{sinx}(0)=1 $

 

[MarkFL]

So
$1+\frac{1}{2}=\frac{3}{2}$

Since

$L_{sinx}(0)=1 $

No, we find:

\(\displaystyle L_{\sin(x)}(0)=x\)

and:

\(\displaystyle L_{\sqrt{x+1}}(0)=\frac{1}{2}x+1\)

Hence:

\(\displaystyle L_{f}(0)=x+\frac{1}{2}x+1=\frac{3}{2}x+1\)

 

[karush]

OK, I was just adding slopes.