Answer
$(a)$
$(i) 2$
$(ii)1.111111$
$(iii)1.010101$
$(iv) 1.001001$
$(v)0.666667$
$(vi)0.909091$
$(vii)0.990099$
$(viii)0.999001$
$(b)1$
$(c) y=x-3$
Work Step by Step
$(a) $
The question requires the slope of the secant line $PQ$. We already know the point $P$($2$, $-1$) from the topic description. Now we just need to find the point $Q$ according to the given condition:
$"$$Q$ is the point ($x$, $\frac{1}{1-x}$ ) $"$
$(i)$ Because the question said $x=1.5$, we know that $Q$ is the point
($1.5$ ,$\frac{1}{1-1.5}$) $=$ ($1.5$ ,$\frac{1}{-0.5}$) $=$ ($1.5$, $-2$)
according to the definition of the slope:
the slope of $PQ=\frac{-1-(-2)}{2-1.5}=\frac{1}{0.5}=2$
$(ii)$ Again, we use the same method to find slope. Because the question said $x=1.5$, $Q$ is the point
($1.9$ ,$\frac{1}{1-1.9}$) $=$ ($1.9$ ,$\frac{1}{-0.9}$) $\approx$ ($1.9$ ,$-1.1111111$)
the slope of $PQ\approx\frac{-1-(-1.1111111)}{2-1.9}=\frac{0.11111111}{0.1}=1.111111$
$(iii)$ As $x=1.99$, $Q$ is the point $(1.99, \frac{1}{1-1.99})=(1.99, -1.01010101)$
the slope of $PQ\approx \frac{-1-(-1.01010101)}{2-1.99}= 1.010101$
$(iv)$ As $x=1.999$, $Q$ is the point
$(1.999, \frac{1}{1-1.999})=(1.999, -1.001001001)$
the slope of $PQ\approx \frac{-1-(-1.001001001)}{2-1.999}= 1.001001$
$(v)$ As $x=2.5$, $Q$ is the point
$(2.5, \frac{1}{1-2.5})=(2.5, -0.666666)$
the slope of $PQ\approx \frac{(-0.666666)-(-1)}{2.5-2}\approx 0.666667$
$(vi)$ As $x=2.1$, $Q$ is the point
$(2.1, \frac{1}{1-2.1})=(2.1, -0.909090)$
the slope of $PQ\approx \frac{(-0.909090)-(-1)}{2.1-2}\approx 0.909091$
$(vii)$ As $x=2.01$, $Q$ is the point
$(2.01, \frac{1}{1-2.01})=(2.01, -0.99009900)$
the slope of $PQ\approx \frac{(-0.99009900)-(-1)}{2.01-2}\approx 0.990099 $
$(viii)$ As $x=2.001$, $Q$ is the point
$(2.001, \frac{1}{1-2.001})=(2.001, -0.999000999)$
the slope of $PQ\approx \frac{(-0.999000999)-(-1)}{2.001-2}\approx 0.999001 $
(b) We can observe the results of part (a). As we let $x$ get closer to 2, we get a slope closer to 1. So we guess that the value of the slope of the tangent line to the curve at $(2, -1)$ is 1.
(c) Using the point slope formula, we can find an equation of the tangent line to the curve at (2, -1).
$1=\frac{y-2}{x-(-1)}$
$y=x-3$
