Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,459 questions

17,854 answers

1,446 comments

805,809 users

Find the linearization L(x) of the function

0 votes

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 20, 2015 in CALCULUS by anonymous
reshown Mar 20, 2015 by bradely

2 Answers

0 votes

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

Related questions

asked Nov 4, 2014 in PRECALCULUS by anonymous
...