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

MATH102 December 2014

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[hide] 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 $g(x)?$ $y=c+{\tfrac {1}{b}}(x-a)$ $y=a+{\tfrac {1}{b}}(x-c)$ $y={\tfrac {1}{a}}+{\tfrac {1}{b}}\left(x-{\tfrac {1}{c}}\right)$ $y=c+b\,(x-a)$ $y=c-{\tfrac {1}{b}}(x-a)$

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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.

[show] 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 $(c,f(c))$ ...)

[show] 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 $f(x)$ )?

[show] 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 $g'(a)?$

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[show] 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 $[f^{-1}]'(x)={\frac {1}{f'(f^{-1}(x))}}.$

[show] 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 ${\color {blue}{\text{(A) }}}y=c+{\tfrac {1}{b}}(x-a).$