Let y = 5x² + 5x + 2. If Ax estimate Ay Δy ~ = 0.2 at x = 5, use linear approximation to

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Chapter1: Functions And Models
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**Problem Statement:**

Let \( y = 5x^2 + 5x + 2 \). If \(\Delta x = 0.2\) at \(x = 5\), use linear approximation to estimate \(\Delta y\).

**Solution:**

To estimate \(\Delta y\) using linear approximation, we need to find the derivative \(\frac{dy}{dx}\) and use it to approximate the change in \(y\) for this small \(\Delta x\).

1. **Find the derivative:**

   \[
   \frac{dy}{dx} = \frac{d}{dx}(5x^2 + 5x + 2) = 10x + 5
   \]

2. **Evaluate the derivative at \(x = 5\):**

   \[
   \frac{dy}{dx} \bigg|_{x=5} = 10(5) + 5 = 50 + 5 = 55
   \]

3. **Estimate \(\Delta y\):**

   \[
   \Delta y \approx \frac{dy}{dx} \bigg|_{x=5} \cdot \Delta x = 55 \cdot 0.2 = 11
   \]

Thus, the estimated change in \(y\) is:

\[ \Delta y \approx 11 \]
Transcribed Image Text:**Problem Statement:** Let \( y = 5x^2 + 5x + 2 \). If \(\Delta x = 0.2\) at \(x = 5\), use linear approximation to estimate \(\Delta y\). **Solution:** To estimate \(\Delta y\) using linear approximation, we need to find the derivative \(\frac{dy}{dx}\) and use it to approximate the change in \(y\) for this small \(\Delta x\). 1. **Find the derivative:** \[ \frac{dy}{dx} = \frac{d}{dx}(5x^2 + 5x + 2) = 10x + 5 \] 2. **Evaluate the derivative at \(x = 5\):** \[ \frac{dy}{dx} \bigg|_{x=5} = 10(5) + 5 = 50 + 5 = 55 \] 3. **Estimate \(\Delta y\):** \[ \Delta y \approx \frac{dy}{dx} \bigg|_{x=5} \cdot \Delta x = 55 \cdot 0.2 = 11 \] Thus, the estimated change in \(y\) is: \[ \Delta y \approx 11 \]
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