MHB Monica's question at Yahoo Answers regarding Linear Approximation

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The discussion centers on finding the linear approximation of the function g(x) = fifth root of (1 + x) at a = 0. The approximation formula is derived using the derivative, resulting in g(x + Δx) ≈ (1/5)Δx + g(x). For Δx = -0.05, the approximation for fifth root of 0.95 is calculated as approximately 0.99, while for Δx = 0.1, the approximation for fifth root of 1.1 is about 1.02. This method effectively demonstrates how linear approximation can be applied to estimate values of functions near a given point. The calculations provide a practical example of using linear approximation in real-world scenarios.
MarkFL
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Here is the question:

Linear Approximation?

Find the linear approximation of the function
g(x) = fifth sqrt(1 + x) at a = 0.

Use it to approximate the numbers
fifth sqrt(0.95) and fifth sqrt(1.1)

I have posted a link there to this thread so the OP can view my work.
 
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Hello Monica,

Consider the approximation:

$$\frac{\Delta g}{\Delta x}\approx\frac{dg}{dx}$$

Now this implies:

$$\Delta g\approx\frac{dg}{dx}\Delta x$$

We may rewrite $\Delta g$ as follows:

$$g\left(x+\Delta x \right)-g(x)\approx\frac{dg}{dx}\Delta x$$

And so we have:

$$g\left(x+\Delta x \right)\approx\frac{dg}{dx}\Delta x+g(x)$$

Now, with $g$ defined as:

$$g(x)\equiv x^{\frac{1}{5}}\implies \frac{dg}{dx}=\frac{1}{5}x^{-\frac{4}{5}}$$

And with $x=1$, our formula becomes:

$$g\left(1+\Delta x \right)\approx\frac{1}{5}\Delta x+1$$

And so for:

i) $$\Delta x=-0.05$$

We have:

$$g\left(1-0.05 \right)\approx\frac{-0.05}{5}+1$$

$$\sqrt[5]{0.95}\approx0.99$$

ii) $$\Delta x=0.1$$

We have:

$$g\left(1+0.1 \right)\approx\frac{0.1}{5}+1$$

$$\sqrt[5]{1.1}\approx1.02$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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