MHB Monica's question at Yahoo Answers regarding Linear Approximation

AI Thread Summary
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.
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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:

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

$$\sqrt[5]{1.1}\approx1.02$$
 
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