Science:Math Exam Resources/Courses/MATH110/December 2014/Question 10/Solution 1

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

$\color {blue}y=-x-1,y=-x+3$