Science:Math Exam Resources/Courses/MATH102/December 2011/Question 01 (b) iii

MATH102 December 2011

  •  Q1 (a) i  •  Q1 (a) ii  •  Q1 (b) i  •  Q1 (b) ii  •  Q1 (b) iii  •  Q1 (c)  •  Q2 (a) i  •  Q2 (a) ii  •  Q2 (a) iii  •  Q2 (b) i  •  Q2 (b) ii  •  Q2 (b) iii  •  Q2 (c)  •  Q3  •  Q4 (i)  •  Q4 (ii)  •  Q4 (iii)  •  Q4 (iv)  •  Q4 (v)  •  Q4 (vi)  •  Q5  •  Q6  •  Q7 (i)  •  Q7 (ii)  •  Q7 (iii)  •  Q8 (i)  •  Q8 (ii)  •  Q8 (iii)  •  Q9  •  

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Question 01 (b) iii
Short-Answer Questions. A correct answer in the box gives full marks. For partial marks work needs to be shown. Let $f^{-1}\,$ be the inverse function of $f(x)\,$. Assume $f(0)=1\,$ and $f'(0)=2.\,$ Find the tangent line $y=mx+b\,$ to $f^{-1}(x)\,$ at 1.

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Hint
The standard form of a line is given by $(y-y_{1})=m(x-x_{1}).$ If the above line lies tangent to the function $f^{-1}(x)$ at $x=1$, what is $m$? If the above line must touch the function $f^{-1}(x)$ at a point, what are the values of $(x_{1},y_{1})$?

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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. To determine the equation of the tangent line to $y=f^{-1}(x)$ at the point $x=1$ we need two things: (1) the slope of $f^{-1}(x)$ at $x=1$ and (2) the value of a point that the tangent line crosses through. (1) Recall this important property of inverse functions: $y=f^{-1}(x)\quad \Leftrightarrow \quad x=f(y)$ From this rule, we can use implicit differentiation to get the derivative for $f^{-1}$ in terms of $f,\,f'$. ${\begin{aligned}{\frac {d}{dx}}\left(x\right)&={\frac {d}{dx}}\left(f(y)\right)\\1&=f'(y){\frac {dy}{dx}}\quad \rightarrow \quad {\frac {dy}{dx}}={\frac {1}{f'(f^{-1}(x))}}\end{aligned}}$ Thus, the slope of the inverse function at the point $x=1$ is given by $m$ where $m={\frac {1}{f'(f^{-1}(1))}}.$ We know that the tangent line must touch the inverse function at the point $x=1$. Using the rules of inverse functions, since $f(0)=1$, we know that $f^{-1}(1)=0$, and thus $m={\frac {1}{f'(f^{-1}(1))}}={\frac {1}{f'(0)}}={\frac {1}{2}}.$ (2) We get by the above point as well: $f^{-1}(1)=0\rightarrow (x_{1},y_{1})=(1,0)$. Using these two things in the standard equation of a line, we get an equation for the tangent line ${\begin{aligned}(y-y_{1})&=m(x-x_{1})\\(y-0)&={\frac {1}{2}}(x-1).\end{aligned}}$ Rearranging the equation to slope-intercept form (as the question asks), we get our final answer: ${\color {blue}y={\frac {x}{2}}-{\frac {1}{2}}}.$

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