Science:Math Exam Resources/Courses/MATH110/December 2013/Question 02

MATH110 December 2013

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q2  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q7  •  Q8  •  Q9  •  Q10 (a)  •  Q10 (b)  •  QS01 10(a)  •  QS01 10(b)  •  

   Other MATH110 Exams

  •  December 2011  •  December 2010  •  April 2012  •  April 2011  •  December 2012  •  April 2013  •  December 2013  •  April 2014  •  December 2014  •  December 2015  •  April 2017  •  April 2016  •  December 2016  •  December 2017  •  April 2018  •  

Question 02
Find the equations of the two tangent lines to the curve $f(x)={\frac {x-4}{x+4}}$ that are parallel to the line x - 2y = 2.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Remember that the derivative of a function at a point is equal to the slope of the tangent line at that point. First find the slope of the given line by changing this equation to the form $\displaystyle y=mx+b$

Hint 2
Once you find the slope of the given line, set it equal to the derivative of the given function, which we find using the quotient rule.

Hint 3
Following the first two hints, we should now have two x-values. Plug these values back into the function definition to find the y-values. After this, you now have the points and the slope of the two lines you need to find to solve the question.

Checking a solution serves two purposes: helping you if, after having used all the hints, 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. We first recognise that parallel means that the slopes must be equal. Rearranging the equation, we get: ${\begin{aligned}x-2y&=2\\x-2&=2y\\y&={\frac {x-2}{2}}\\y&={\frac {1}{2}}x-1.\end{aligned}}$ So that means we are looking for points along the curve with slope equal to ${\frac {1}{2}}$. To get those points, we need the derivative of the curve: ${\begin{aligned}{\frac {\rm {d}}{{\rm {d}}x}}(y)&={\frac {\rm {d}}{{\rm {d}}x}}\left({\frac {x-4}{x+4}}\right)\end{aligned}}$ Using the quotient rule, we get: ${\begin{aligned}u(x)&=x-4&u'(x)=1\\v(x)&=x+4&v'(x)=1\end{aligned}}$ So that means: ${\begin{aligned}{\frac {{\rm {d}}{y}}{{\rm {d}}x}}&={\frac {v(x)u'(x)-u(x)v'(x)}{\left(v(x)\right)^{2}}}\\&={\frac {(x+4)\cdot (1)-(x-4)\cdot (1)}{\left(x+4\right)^{2}}}\\&={\frac {8}{\left(x+4\right)^{2}}}\end{aligned}}$ We are interested in points where the derivative is ${\frac {1}{2}}$ ${\begin{aligned}{\frac {1}{2}}&={\frac {8}{\left(x+4\right)^{2}}}\\\left(x+4\right)^{2}&=16\\x+4&=\pm 4\\x&=0,-8\end{aligned}}$ This means we have to tackle two potential $x$ values. At $x=0$, we have: ${\begin{aligned}y&={\frac {0-4}{0+4}}=-1\end{aligned}}$ At $x=-8$, we have: ${\begin{aligned}y&={\frac {-8-4}{-8+4}}={\frac {-12}{-4}}=3\end{aligned}}$ So that means the tangent lines are are looking for are: for $x=0$. ${\begin{aligned}y&={\frac {1}{2}}(x-0)+(-1)\\&={\frac {1}{2}}x-1\\\end{aligned}}$ and for $x=-8$ ${\begin{aligned}y&={\frac {1}{2}}(x-(-8))+3\\&={\frac {1}{2}}(x+8)+3\\&={\frac {1}{2}}x+7\\\end{aligned}}$

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER QGT flag, MER Tag Quotient rule, MER Tag Tangent line, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag