Science:Math Exam Resources/Courses/MATH102/December 2014/Question A 01

MATH102 December 2014

Work in progress: this question page is incomplete, there might be mistakes in the material you are seeing here.

  •  QA 1  •  QA 2  •  QA 3  •  QA 4  •  QA 5  •  QA 6  •  QA 7  •  QA 8  •  QB 1  •  QB 2  •  QB 3  •  QB 4  •  QB 5  •  QB 6  •  QB 7  •  QC 1  •  QC 2(a)  •  QC 2(b)  •  QC 2(c)  •  QC 2(d)  •  QC 3  •  

   Other MATH102 Exams

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

Question A 01
Suppose that $g(x)$ is the inverse function of $f(x).$ The tangent line to $f(x)$ at $x=c$ is $y=a+b\,(x-c)$ where $a,b$ and $c$ are constants. Which of the following is the equation of a tangent line to $g(x)?$ $y=c+{\tfrac {1}{b}}(x-a)$ $y=a+{\tfrac {1}{b}}(x-c)$ $y={\tfrac {1}{a}}+{\tfrac {1}{b}}\left(x-{\tfrac {1}{c}}\right)$ $y=c+b\,(x-a)$ $y=c-{\tfrac {1}{b}}(x-a)$

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
Recall that the tangent line to $f(x)$ at $x=c$ is given by $y=f(c)+f'(c)(x-c).$ (Why? Consider the slope of the tangent line and the fact that it must pass through $(c,f(c))$ ...)

Hint 2
Following the first hint, we see that $a=f(c)$ and $b=f'(c).$ Given this information, what point must lie on the graph of $g(x)$ (the inverse of $f(x)$)?

Hint 3
$a=f(c)$ implies that $g(a)=c$ since $g=f^{-1}.$ Hence the point $(a,c)$ must lie on the graph of $g(x).$ What is $g'(a)?$

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 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. As the tangent line to $f(x)$ at $x=c$ is given by $y={\color {OliveGreen}f(c)}+{\color {OrangeRed}f'(c)}(x-c),$ it is clear that ${\color {OliveGreen}a}=f(c)$ and ${\color {OrangeRed}b}=f'(c).$ Since $g=f^{-1},\ a=f(c)\implies {\color {Orange}g(a)}=c.$ Now recall* that $[f^{-1}]'(x)={\frac {1}{f'(f^{-1}(x))}},$ so $g'(a)=[f^{-1}]'(a)={\frac {1}{f'(g(a))}}={\frac {1}{\color {OrangeRed}f'(c)}}={\frac {1}{b}}.$ Since the tangent line to $g(x)$ at $x=a$ is given by $y={\color {Orange}g(a)}+g'(a)(x-a),$ the line ${\color {blue}{\text{(A) }}}y=c+{\tfrac {1}{b}}(x-a)$ is tangent to the graph of $g(x)$ as desired. * If you do not recall this, observe that by the chain rule, $[f(f^{-1}(x))]'=f'(f^{-1}(x))\cdot [f^{-1}]'(x).$ But $f(f^{-1}(x))=x,$ and $x'=1,$ so $[f^{-1}]'(x)={\frac {1}{f'(f^{-1}(x))}}.$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Since the graph of $g(x)$, the inverse of $f(x)$, is simply the graph of $f(x)$ reflected across the line $y=x,$ a tangent line to the graph of $g(x)$ can be found by simply reflecting a tangent line to the graph of $f(x)$ across the same line. In other words, 'a tangent to the inverse is the inverse of a tangent'. Algebraically, we simply interchange $x$ and $y$ in the equation $y=a+b\,(x-c),$ thereby obtaining $x=a+b\,(y-c).$ Solving for $y,$ we arrive at ${\color {blue}{\text{(A) }}}y=c+{\tfrac {1}{b}}(x-a).$

Click here for similar questions

MER QGQ flag, MER QRH flag, MER RS flag, MER RT flag, MER Tag Tangent line, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag