Science:Math Exam Resources/Courses/MATH104/December 2011/Question 01 (d)

MATH104 December 2011

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q1 (i)  •  Q1 (j)  •  Q1 (k)  •  Q1 (l)  •  Q1 (m)  •  Q1 (n)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q2 (e)  •  Q2 (f)  •  Q2 (g)  •  Q3  •  Q4  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q5 (d)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q6 (d)  •  Q6 (e)  •  

   Other MATH104 Exams

  •  December 2014  •  December 2011  •  December 2010  •  December 2012  •  December 2013  •  December 2015  •  December 2016  •  

Question 01 (d)
Find the equation of the tangent line to $y=f(x)={\frac {x+3}{2x+1}}$ at the point corresponding to x = 0.

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
In order to find the equation of any line (including a tangent), you need to know the slope of the line and the coordinates of a point on the line. How can you find the slope of a tangent line? What is the point on $f(x)$ (and thus the tangent line as well) at $x=0$?

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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 following solution makes use of the product rule, rather than the quotient rule, for differentiation. Writing the given function as: $\displaystyle y(x)=(x+3)(2x+1)^{-1}$ allows us to apply the product rule when taking the derivative: ${\frac {dy}{dx}}=y'(x)={\frac {1}{2x+1}}-{\frac {2x+6}{(2x+1)^{2}}}$ At x = 0, the derivative of y is equal to ${\begin{aligned}y'(0)&={\frac {1}{2(0)+1}}-{\frac {2(0)+6}{(2(0)+1)^{2}}}\\&=1-6\\&=-5\end{aligned}}$ The general "point-slope form" for the equation of a straight line is $\displaystyle y-y_{1}=m(x-x_{1})$ where (x1,y1) is a point on the line with slope m. We choose x1 = 0 and hence y1 = y(0) = 3. Plugging in these values yields ${\begin{aligned}y-y(0)&=(-5)(x-0)\\y-3&=-5x\\y&=-5x+3\end{aligned}}$ Therefore, the required equation is y = -5x +3.

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Tangent line, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag