Find the linear approximation of f(x)=\ln x at x=1 and use it to estimate ln 1.12.
L(x)= .
ln 1.12 approx
i got my L(x) = x-1 but for some reason i can't seem to get the second part of the answer with ln 1.12 approximation
To find the linear approximation of \(f(x)=\ln x\) at \(x=1\), we can use the formula for the tangent line approximation:
\[L(x) = f(a) + f'(a)(x-a)\]
where \(a\) is the value at which we want to approximate the function, and \(f'(x)\) is the derivative of the function.
In this case, we want to approximate \(f(x)=\ln x\) at \(x=1\). So our formula becomes:
\[L(x) = \ln 1 + \frac{d}{dx}(\ln x)(x-1)\]
Now, let's find the derivative of \(\ln x\):
\[\frac{d}{dx}(\ln x) = \frac{1}{x}\]
Substituting this back into our formula, we get:
\[L(x) = \ln 1 + \frac{1}{1}(x-1)\]
Simplifying, we have:
\[L(x) = 0 + (x-1)\]
So the linear approximation is simply \(L(x) = x-1\).
To estimate \(\ln 1.12\) using this linear approximation, we substitute \(x=1.12\) into the linear approximation:
\[L(1.12) = 1.12 - 1 = 0.12\]
Therefore, the estimated value of \(\ln 1.12\) using the linear approximation is \(0.12\).