Science:Math Exam Resources/Courses/MATH110/December 2014/Question 10

MATH110 December 2014

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Question 10
Find equations of all lines having slope -1 that are tangent to the curve ${\textstyle \displaystyle y={\frac {1}{x-1}}}$.

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
Recall that the formula for the tangent line of a curve $y=f(x)$ at a point $a$ is given by $y=f'(a)(x-a)+f(a)$

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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. Recall that the formula for the tangent line of a curve $y=f(x)$ at a point $a$ is given by $y=f'(a)(x-a)+f(a)$. This is, indeed, a line with slope $f'(a)$ passing though $(a,f(a))$. Since we are looking for the tangent line(s) with slope $-1$, first we solve $f'(a)=-1$. By the chain rule with $u=x-1$, the derivative of $f$ can be obtained by $f'(x)={\frac {df}{dx}}={\frac {df}{du}}\cdot {\frac {du}{dx}}={\frac {d}{du}}{\frac {1}{u}}\cdot {\frac {d}{dx}}(x-1)=-{\frac {1}{u^{2}}}=-{\frac {1}{(x-1)^{2}}}$. Then, $f'(a)=-1$ if and only if $-{\frac {1}{(a-1)^{2}}}=-1\iff (a-1)^{2}=1\iff a-1=\pm 1\iff a=0,{\text{ or }}a=2$. In the case of $a=0$, we get $f(0)={\frac {1}{0-1}}=-1$, therefore the corresponding tangent line is $y=f'(0)(x-0)+f(0)=-x-1$. On the other hand, for $a=2$, we have $f(2)={\frac {1}{2-1}}=1$, so that $y=f'(2)(x-2)+f(2)=-1(x-2)+1=-x+3$. Therefore, the tangent lines of the given function $f(x)$ with slope $-1$ are $\color {blue}y=-x-1,y=-x+3$