Question

exercises 4.2 linear approximations and differentials
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question 6
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let $y = 3x^2 + 3x + 4$. if $\delta x = 0.3$ at $x = 4$, use linear approximation to estimate $\delta y$
$\delta y \approx $
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Explanation:

Step1: Recall Linear Approximation Formula

The linear approximation for \(\Delta y\) is given by \(\Delta y \approx dy = y'(x) \cdot \Delta x\), where \(y'\) is the derivative of \(y\) with respect to \(x\).

Step2: Find the Derivative of \(y\)

Given \(y = 3x^2 + 3x + 4\), we find its derivative using the power rule. The derivative \(y' = \frac{d}{dx}(3x^2 + 3x + 4)= 6x + 3\).

Step3: Evaluate the Derivative at \(x = 4\)

Substitute \(x = 4\) into \(y'\): \(y'(4)= 6(4)+ 3 = 24 + 3 = 27\).

Step4: Calculate \(\Delta y\) Approximation

We know \(\Delta x = 0.3\) and \(y'(4)= 27\). Using \(\Delta y \approx y'(x)\cdot\Delta x\), we substitute the values: \(\Delta y \approx 27\times0.3 = 8.1\).

Answer:

\(8.1\)