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Find the linearization L(x) of the function f(x) = ln(1 + x)

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Find the linearization L(x) of the function f(x) = ln(1 + x) at a = 0 and use it to approximate the numbers ln 1.2 and ln 1.01. Compare the estimates from the linear approximation with the values given by a calculator. Which one of the two estimates is more accurate, the estimation of ln 1.2 or the estimation of ln 1.01?
asked Mar 15, 2015 in CALCULUS by anonymous

2 Answers

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Step 1:

The function is f(x)=\ln \ (1+x).

The linear approximation of a function is L(x)=f(a)+f'(a)(x-a).

Find the derivative of the function at x=a.

\\ \frac{d}{dx}f(x)=\frac{d}{dx}\left (\ln \ (1+x) \right )\\ \\ \\f'(x)=\frac{1}{1+x}\ \frac{d}{dx} \left ( 1+x \right )\\ \\ \\f'(x)=\frac{1}{1+x}\ (0+1) \\ \\ \\f'(x)=\frac{1}{1+x}\ \\ \\ \\f'(a)=\frac{1}{1+a}

Now put x=a in f(x)=\ln \ (1+x).

f(a)=\ln \ (1+a)

Step 2:

L(x)=f(a)+f'(a)(x-a)

L(x)=\ln\ (1+a)+\left ( \frac{1}{1+a} \right )\left ( x-a \right )

Now find the  linear approximation at a=0.

\\L(x)=\ln\ (1+0)+\left ( \frac{1}{1+0} \right )\left ( x-0 \right )\\ \\L(x)=\ln\ (1)+\left ( 1 \right )\left ( x \right )\\ \\L(x)=0+ ( x)\\ \\L(x)=x

So the linear approximation at a=0 is  \\L(x)=x.

answered Mar 20, 2015 by yamin_math Mentor
0 votes

Contd...

Step 3:

Now use the linear approximation to find the  \ln \ (1.2) and \ln \ (1.01).

Rewrite  \ln \ (1.2) = \ln \ (1+0.2).

Now compare it with the function \ln \ (1+x).

Here x=0.2.

Now approximate for x=0.2.

\\L(x)=x \\ \\L(0.2)=0.2

Step 4:

Rewrite  \ln \ (1.01) = \ln \ (1+0.01).

Now compare it with the function \ln \ (1+x).

Here x=0.01.

Now approximate for x=0.01.

\\L(x)=x \\ \\L(0.01)=0.01

Step 5:

Now calculate \ln \ (1.2) and \ln \ (1.01) using calculator.

\ln \ (1.2)=0.18232

\ln \ (1.01)=0.00995

The second approximation is significantly better, since it is significantly closer to 0 than 1.2.

Solution :

(1) The linear approximation is \\L(x)=x.

(2) \\L(0.2)=0.2.

(3) \\L(0.01)=0.01.

(4) The second approximation is more accurate.

answered Mar 20, 2015 by yamin_math Mentor

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