Question

linearization and differentials: problem 1
(1 point)
find the linear approximation of ( f(x)=ln x ) at ( x = 1 ) and use it to estimate ( ln (1.19) ).
( l(x)=)
( ln 1.19 approx )
note: you can earn partial credit on this problem

Explanation:

Step1: Find the derivative of \(f(x)\)

The function \(f(x)=\ln x\), its derivative \(f^\prime(x)=\frac{1}{x}\).

Step2: Evaluate \(f(x)\) and \(f^\prime(x)\) at \(x = 1\)

When \(x = 1\), \(f(1)=\ln(1)=0\), and \(f^\prime(1)=\frac{1}{1}=1\).

Step3: Use the linear - approximation formula \(L(x)=f(a)+f^\prime(a)(x - a)\)

Here \(a = 1\), so \(L(x)=f(1)+f^\prime(1)(x - 1)\). Substituting \(f(1)=0\) and \(f^\prime(1)=1\) into the formula, we get \(L(x)=0 + 1\times(x - 1)=x-1\).

Step4: Estimate \(\ln(1.19)\) using the linear approximation

Let \(x = 1.19\) in \(L(x)\). Then \(\ln(1.19)\approx L(1.19)\). Since \(L(x)=x - 1\), when \(x = 1.19\), \(L(1.19)=1.19-1=0.19\).

Answer:

\(L(x)=x - 1\); \(\ln(1.19)\approx0.19\)