Science:Math Exam Resources/Courses/MATH104/December 2013/Question 01 (i)

MATH104 December 2013

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Question 01 (i)
Find an approximation to $\displaystyle \ln(1.25)$ by using the linear approximation to $\displaystyle f(x)=\ln x$ at $\displaystyle a=1$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that a linear approximation involves finding the equation of the tangent line to $\displaystyle f$ at the point $a=1$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We use the point slope formula ${\frac {y-f(a)}{x-a}}=f'(a).$ Since $a=1$ we obtain $f(a)=\ln(1)=0$, while $f'(x)={\frac {1}{x}}$ implies $f'(a)=1$. Hence $\displaystyle {\frac {y-0}{x-1}}=1.$ Solving gives $\displaystyle y=x-1$ Thus, the approximation we seek is given by $\displaystyle y=1.25-1=0.25$.

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