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I need to find the tangent line at $(-1, f(-1))$ of the following equation: $f(x) = \dfrac{x+2}{1-x}-3x$

I tried this: $ \lim_{h \to 0} \dfrac{f(-1+h) - f(-1)}{h}$ which is giving me this: $\lim_{h \to 0} \dfrac{\dfrac{-1+h+2}{1-1+h} -3(-1+h)-\dfrac{-1+2}{1--1} -3(-1)}{h}=\lim_{h \to 0} \dfrac{\dfrac{h+1}{h} -3h - 0.5}{h}$ But now, I'm stuck, I can't replace $h$ with $0$ and simplify.

Any help would be appreciated.

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    There was a sign error when you substituted. It is not $\frac{-1+h+2}{1-1+h}$. The denominator should be $1-(-1+h)=2-h$. I suggest you use more parentheses.2012-11-24

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The slope of the tangent at $(-1,f(-1))$ can be found by differentiating $f(x)$ and plugging in $x=-1$. Based on your work, it looks like want to evaluate the slope using first principles. Hence, slope at $x=-1$ is given by \begin{align} \lim_{h \to 0} \dfrac{f(-1+h)-f(-1)}h & = \lim_{h \to 0} \dfrac{\left(\dfrac{-1+h+2}{1-(-1+h)}-3(-1+h) \right) - \left(\dfrac{-1+2}{1-(-1)}-3(-1) \right)}h\\ & = \lim_{h \to 0} \dfrac{\left(\dfrac{h+1}{2-h}+3 - 3h \right) - \left(\dfrac{1}{2}+3 \right)}h = \lim_{h \to 0} \dfrac{\left(\dfrac{h+1}{2-h} - 3h \right) - \dfrac{1}{2}}h\\ & = \lim_{h \to 0} \dfrac{\overbrace{\left(\dfrac{h+1}{2-h} - \dfrac{1}{2} \right)}^{\text{Take the l.c.m}} - 3h}h\\ & = \lim_{h \to 0} \dfrac{\left(\dfrac{2h+2-2+h}{4-2h} \right)-3h}h = \lim_{h \to 0} \dfrac{\left(\dfrac{3h}{4-2h} \right)-3h}h\\ & = \lim_{h \to 0} \left(\dfrac3{4-2h} - 3 \right) = \dfrac34 -3 = -\dfrac94 \end{align} Hence, the slope of the tangent at $(-1,f(-1))$ is $-9/4$. You also know that the tangent passes through $(-1,f(-1))$. Note that $f(-1) = \left(\dfrac{-1+2}{1-(-1)}-3(-1) \right) = \dfrac72$ Hence, the tangent passes through $(-1,7/2)$ and has slope $-9/4$. The equation of the tangent is $\dfrac{y-7/2}{x+1} = \dfrac{-9}4 \implies y = \dfrac72 - \dfrac94(x+1) = -\dfrac94x + \dfrac54$ Hence, the equation is $4y+9x=5$

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    This is very cle$a$r, thank you very much, it's really appreciated !2012-11-24