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How do you use the tangent line approximation to approximate the value of #ln(1.006)# ?

Answer 1

Emilia Aguilar

#ln(1.006) approx 0.006#

Let us look at some details.

Let #f(x)=lnx#. #Rightarrow f'(x)=1/x#

#f(1)=ln1=0# #f'(1)=1/1=1#

So, the tangent line approximation #L(x)# can be written as

#L(x)=f(1)+f'(1)(x-1)=0+1(x-1)=x-1#

Hence,

#f(1.006) approx L(1.006)=1.006-1=0.006#

I hope that this was helpful.

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Answer 2

Christopher Allers

To approximate the value of ln(1.006) using the tangent line approximation, you would first find a point near 1.006 where you know the natural logarithm. Let's choose 1 as it's easy to calculate. Then, you would calculate the derivative of ln(x) at x = 1, which is simply 1.

The tangent line approximation formula is:

f(x) ≈ f(a) + f'(a)(x - a)

Plugging in the values: f(1.006) ≈ ln(1) + 1(1.006 - 1)

Simplify and calculate: f(1.006) ≈ 0 + 1(0.006) f(1.006) ≈ 0.006

So, the approximation of ln(1.006) using the tangent line at x = 1 is approximately 0.006.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.