Science:Math Exam Resources/Courses/MATH102/December 2019/Question 05 (a)

MATH102 December 2019

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Question 05 (a)
For this question, let $f(x)=1-(1+x)^{-1}$ You may use the following facts without proving them: $f(x)={\frac {x}{1+x}}$ $f(x)={\frac {1}{(1+x)^{2}}}$ $f(x)={\frac {-2}{(1+x)^{3}}}$ (a) Give an equation of the tangent line to $y=f(x)$ at the point $\left(10,{\frac {10}{11}}\right)$.

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
For a curve described by a function, at any point, the slope of the tangent line is equal to the derivative of the function.

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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. The slope of the tangent line to the curve $y=f(x)$ is equal to $f'(x)$. $f'(10)={\frac {1}{(1+10)^{2}}}={\frac {1}{121}}$. Using the slope-point form, the tangent line equation is $y-{\frac {10}{11}}={\frac {1}{121}}\left(x-10\right)$, i.e. $y={\frac {1}{121}}x+{\frac {100}{121}}$.