Linearizing Sqrt(x) At X=1/9 For Approximations

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Hey everyone, let's dive into the awesome world of calculus and tackle a cool problem involving linearization. We're going to find the linearization of the function $f(x) = \sqrt{x}$ at the point $x = \frac{1}{9}$. Then, we'll use this linearization, which we'll call $L(x)$, to approximate the value of $\sqrt{\frac{1}{8}}$. This is a fantastic way to see how we can use tangent lines to estimate values of functions that might be tricky to calculate directly. So, buckle up, grab your favorite thinking cap, and let's get this math party started!

Understanding Linearization: Your New Best Friend for Approximations

Alright guys, let's talk about what linearization actually is. Basically, linearization is a fancy term for finding the equation of the tangent line to a function at a specific point. Why do we care about this tangent line? Because, near that point, the tangent line is a super-close approximation of the original function. Think of it like this: if you zoom in really, really close to a point on a curve, it starts to look like a straight line. That straight line is precisely what our linearization captures! The formula for linearization, or the tangent line approximation, of a function $f(x)$ at a point $a$ is given by:

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

Here's the breakdown of what each part means:

• $f(a)$: This is the value of the function at the point $a$. It's our starting point on the curve.
• $f'(a)$: This is the derivative of the function evaluated at $a$. Remember, the derivative gives us the slope of the tangent line at that point. It tells us how steep the function is right at $a$.
• $(x-a)$: This is the difference between any point $x$ and our specific point of tangency $a$. It's how far we're moving horizontally from $a$.

So, the whole equation $L(x) = f(a) + f'(a)(x-a)$ is just the point-slope form of a line, where $f(a)$ is our $y$-coordinate and $f'(a)$ is our slope. It's a powerful tool because it allows us to approximate function values for inputs $x$ that are close to $a$, without having to plug them directly into the original (potentially complicated) function $f(x)$. This was historically super important before calculators and computers became commonplace!

Step-by-Step: Finding the Linearization of $f(x) = \sqrt{x}$ at $x = \frac{1}{9}$

Okay, let's get down to business and find our linearization. Our function is $f(x) = \sqrt{x}$, and we're interested in the point $a = \frac{1}{9}$.

Step 1: Find $f(a)$

First, we need to evaluate our function at $a = \frac{1}{9}$.

$f(\frac{1}{9}) = \sqrt{\frac{1}{9}}$

We know that the square root of $\frac{1}{9}$ is $\frac{1}{3}$.

$f(\frac{1}{9}) = \frac{1}{3}$

Step 2: Find the derivative $f'(x)$

Next, we need to find the derivative of $f(x) = \sqrt{x}$. It's often easier to work with exponents, so let's rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$.

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

Now, we can use the power rule for differentiation, which states that $\frac{d}{dx}(x^n) = nx^{n-1}$.

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

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

We can also rewrite this with a positive exponent and in radical form:

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

Step 3: Find $f'(a)$

Now, we evaluate the derivative at our point $a = \frac{1}{9}$.

$f'(\frac{1}{9}) = \frac{1}{2\sqrt{\frac{1}{9}}}$

We already know that $\sqrt{\frac{1}{9}} = \frac{1}{3}$. So, substitute that in:

$f'(\frac{1}{9}) = \frac{1}{2(\frac{1}{3})}$

$f'(\frac{1}{9}) = \frac{1}{\frac{2}{3}}$

Dividing by a fraction is the same as multiplying by its reciprocal:

$f'(\frac{1}{9}) = 1 \times \frac{3}{2}$

$f'(\frac{1}{9}) = \frac{3}{2}$

Step 4: Write the Linearization $L(x)$

Now we have all the pieces to plug into our linearization formula: $L(x) = f(a) + f'(a)(x-a)$.

We have $a = \frac{1}{9}$, $f(a) = \frac{1}{3}$, and $f'(a) = \frac{3}{2}$.

$L(x) = \frac{1}{3} + \frac{3}{2}(x - \frac{1}{9})$

Let's simplify this a bit:

$L(x) = \frac{1}{3} + \frac{3}{2}x - \frac{3}{2} \times \frac{1}{9}$

$L(x) = \frac{1}{3} + \frac{3}{2}x - \frac{3}{18}$

We can simplify $\frac{3}{18}$ to $\frac{1}{6}$.

$L(x) = \frac{1}{3} + \frac{3}{2}x - \frac{1}{6}$

Now, let's combine the constant terms. To do this, we need a common denominator, which is 6.

$\frac{1}{3} = \frac{2}{6}$

So, $\frac{2}{6} - \frac{1}{6} = \frac{1}{6}$.

$L(x) = \frac{3}{2}x + \frac{1}{6}$

And there you have it! The linearization of $f(x) = \sqrt{x}$ at $x = \frac{1}{9}$ is $L(x) = \frac{3}{2}x + \frac{1}{6}$. This straight line is a great approximation for $\sqrt{x}$ when $x$ is close to $\frac{1}{9}$.

Approximating $\sqrt{\frac{1}{8}}$ Using Linearization

Now for the fun part – using our $L(x)$ to approximate $\sqrt{\frac{1}{8}}$. We want to find the value of $\sqrt{\frac{1}{8}}$. Since $\frac{1}{8}$ is pretty close to our point of linearization, $a = \frac{1}{9}$, we can use $L(x)$ to estimate it.

We need to plug $x = \frac{1}{8}$ into our linearization formula $L(x) = \frac{3}{2}x + \frac{1}{6}$.

$L(\frac{1}{8}) = \frac{3}{2}(\frac{1}{8}) + \frac{1}{6}$

First, multiply the fraction:

$L(\frac{1}{8}) = \frac{3}{16} + \frac{1}{6}$

To add these fractions, we need a common denominator. The least common multiple of 16 and 6 is 48.

Convert $\frac{3}{16}$ to 48ths:

$\frac{3}{16} = \frac{3 \times 3}{16 \times 3} = \frac{9}{48}$

Convert $\frac{1}{6}$ to 48ths:

$\frac{1}{6} = \frac{1 \times 8}{6 \times 8} = \frac{8}{48}$

Now, add them:

$L(\frac{1}{8}) = \frac{9}{48} + \frac{8}{48}$

$L(\frac{1}{8}) = \frac{17}{48}$

So, our linearization tells us that $\sqrt{\frac{1}{8}}$ is approximately $\frac{17}{48}$.

Let's check how good this approximation is. We know that $\sqrt{\frac{1}{8}} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$. Using a calculator, $\sqrt{2} \approx 1.4142$, so $\frac{\sqrt{2}}{4} \approx \frac{1.4142}{4} \approx 0.35355$.

Now let's see what $\frac{17}{48}$ is as a decimal:

$\frac{17}{48} \approx 0.354166...$

Wow, that's a pretty darn close approximation! The difference is quite small, which highlights the power of linearization for estimating values near a known point.

Why Does This Work? The Magic of Tangent Lines

So, why is this linearization thing so effective? It all comes down to the behavior of functions and their tangent lines. Remember, the derivative $f'(a)$ represents the instantaneous rate of change of the function at point $a$. The tangent line at $a$ has exactly this slope. When we are very close to $a$, the curve of the function $f(x)$ and the straight line $L(x)$ are almost indistinguishable. Imagine standing on a hill; if you look just a few feet around you, the ground looks pretty flat, even though the hill might be quite steep overall. The tangent line is like that