Question

How do you find the linearization at $a = 1$ of $f(x) = \sqrt {x + 3} ?$

Answer 1

The linearization of a differentiable function f at a point $x = a$ is the linear function $L(x) = f(a) + f'(a)(x - a)$, whose graph is the tangent line to the graph of $f$ at the point $(a,f(a))$. When $x \approx a$, we get the approximation $f(x) \approx L(x)$.

Formula used:
$L(x) = f(a) + f'(a)(x - a)$
$f'(a) = \dfrac{{df(a)}}{{dx}}$

Complete step by step solution:
For finding the linearization of any differentiable function we use below given equation:
$L(x) = f(a) + f'(a)(x - a)$
In this equation $f(a)$ is given function at ‘a’ and $f'(a)$ is differential of $f(a)$.
So first we will find $f'(x)$
$f'(x) = \dfrac{{df(x)}}{{dx}} = \dfrac{{d\sqrt {x + 3} }}{{dx}} \\\ \\\$
Here, $\sqrt {x + 3} = {(x + 3)^{\dfrac{1}{2}}}$
$\Rightarrow f'(x) = \frac{1}{2}{(x + 3)^{\frac{{ - 1}}{2}}}$
Now we will put $x = 1$,
$f(1) = \sqrt {1 + 3} \\\ = \sqrt 4 \\\ = 2 \\\ f'(1) = \dfrac{1}{2}{(1 + 3)^{\dfrac{{ - 1}}{2}}} \\\ = \dfrac{1}{2}{(4)^{\dfrac{{ - 1}}{2}}} \\\ = \dfrac{1}{2}(\dfrac{1}{2}) \\\ = \dfrac{1}{4} \\\$
So, here $f(1) = 2{\text{ }}and{\text{ }}f'(1) = \dfrac{1}{4}$
After putting values in equation, we get,
$L(x) = f(a) + f'(a)(x - a) \\\ = 2 + \dfrac{1}{4}(x - 1) \\\ = \dfrac{{8 + x - 1}}{4} \\\ = \dfrac{{x + 7}}{4} \\\$

So, here is the right answer to this question, linearization of a differentiable function f at a point $x = a$ is $\dfrac{{x + 7}}{4}$.

Note:
For solving such types of questions first you must have to understand the question and find (try to remember) the best way to solve it. you should first write-down what has been given to you and what you have to find. Then use formulas to solve it.