# Linear Approximation of Single Variable Functions

Suppose that we have a function $f$, and we take a look at the tangent line at point $(a, f(a))$ on $f$. As we zoom in closer and closer towards $(a, f(a))$, we notice that $f$ begins to look closer and closer line the tangent line at $a$ as illustrated in the diagram below:

Sometimes it is very difficult or impossible to calculate values very close to $a$, so we use the graph of the linear tangent line at $a$ as an approximation. Recall that the equation of a tangent line at $(a, f(a))$ comes in the form $y - f(a) = f'(a)(x - a)$. We will now define some important terms.

Definition: If $f$ is a differentiable function and $f'(a)$ exists, then for $x$ very close to $a$ in the domain of $f$, $f(x) \approx f(a) + f'(a)(x - a)$ is known as the Linear Approximation of $f$ at $a$ and the function $L(x) = f(a) + f'(a)(x - a)$ is known as the Linearization of $f$ at $a$. |

*We will see later that the linearization of $f$ at $a$ is actually the first order Taylor polynomial of $f$ at $a$, and we will see that higher order polynomials can be used to approximate values $f(x)$ when $x$ is near $a$.*

Let's look at an example of how we can use a linearization to find a linear approximation of a value of a function.

## Example 1

**Find the linearization of $f(x) = \sqrt{x + 8}$ and at $a = 1$, use it to approximate the values of $\sqrt{8.05}$ and $\sqrt{7.95}$.**

We first must differentiate $f$ to get that $f'(x) = \frac{1}{2\sqrt{x + 8}}$. We note that $f(1) = \sqrt{1 + 8} = 3$, and $f'(1) = \frac{1}{2\sqrt{1 + 8}} = \frac{1}{6}$. Since we know $f(1)$, and $f'(1)$, we substitute these in to get:

(1)Now let $L(x) = \sqrt{8.05}$, and thus:

(2)Similarly, let $L(x) = \sqrt{7.95}$, and therefore:

(3)