Question

1. - / 2 points this is section 3.1 problem 42: for y = f(x)=xe^{x - 5}, when x = 5 and dx = 0.1: dy=, hence the linear approximation using dy is f(5.1)≈f(5)+dy)=. hint: follow example 5. resources ebook

Explanation:

Step1: Differentiate \(y = xe^{x - 5}\) using product - rule

The product - rule states that if \(y=uv\), where \(u = x\) and \(v=e^{x - 5}\), then \(y^\prime=u^\prime v+uv^\prime\). We know that \(u^\prime = 1\) and \(v^\prime=e^{x - 5}\). So \(y^\prime=(1)\times e^{x - 5}+x\times e^{x - 5}=(x + 1)e^{x - 5}\).

Step2: Calculate \(dy\)

We know that \(dy=y^\prime dx\). Substitute \(x = 5\) and \(dx=0.1\) into the formula. When \(x = 5\), \(y^\prime=(5 + 1)e^{5 - 5}=6\times1 = 6\). Then \(dy=y^\prime dx=6\times0.1 = 0.6\).

Step3: Calculate \(f(5)\)

Substitute \(x = 5\) into \(y = xe^{x - 5}\), we get \(f(5)=5\times e^{5 - 5}=5\times1 = 5\).

Step4: Calculate the linear approximation

The linear approximation formula is \(f(x+\Delta x)\approx f(x)+dy\). Here \(x = 5\), \(\Delta x=0.1\), so \(f(5.1)\approx f(5)+dy=5 + 0.6=5.6\).

Answer:

\(dy = 0.6\)
\(f(5.1)\approx5.6\)