Answer
a. (i) = $-2$; (ii) = $-2$
b. $y = -2x +2$
c. Graph; this graph is where the line and the curve coincide.
Work Step by Step
Definition 1: $m = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x-a}$
Since we are asked to evaluate the slope at $(1,0)$, so $a = 1$.
$m = \lim\limits_{x \to 1} \frac{f(x) - f(1)}{x-1}$
Now change $f(x)$ to $(x-x^{3})$.
$m = \lim\limits_{x \to 1} \frac{x-x^{3} - (1-1^{3})}{x-1}$
$m = \lim\limits_{x \to 1} \frac{x-x^{3} - 0}{x-1}$
$m = \lim\limits_{x \to 1} \frac{x-x^{3}}{x-1}$
Simplify:
$m = \lim\limits_{x \to 1} \frac{x(1-x^{2})}{x-1}$
Expand:
$m = \lim\limits_{x \to 1} \frac{x(1-x)(1+x)}{x-1}$
Now multiply the numerator by $-1$.
$m = \lim\limits_{x \to 1} \frac{-x(x-1)(1+x)}{x-1}$
Cancel out: $x-1$.
$m = \lim\limits_{x \to 1} -x(1+x)$
Now replace $x$ to $1$.
$m = -1(1+1) = -2$
(ii) Using Equation 2
$m = \lim\limits_{h \to 0} \frac{f(a+h)-f(a)}{h}$
$m = \lim\limits_{h \to 0} \frac{((a+h)-(a+h)^{3} - (a-a^{3})}{h}$
Now expand $(a+h)^{3}$
$m = \lim\limits_{h \to 0} \frac{(a+h)-(a^{3} +3a^{2}h + 3ah^{2} + h^{3}) - (a-a^{3})}{h}$
Now replace $a$ with $1$.
$m = \lim\limits_{h \to 0} \frac{(1+h)-(1^{3} +3(1)^{2}h + 3(1)h^{2} + h^{3}) - (1-1^{3})}{h}$
$m = \lim\limits_{h \to 0} \frac{-h^{3} - 3h^{2} - 2h}{h}$
Simplify:
$m = \lim\limits_{h \to 0} \frac{h(-h^{2} - 3h - 2)}{h}$
Cancel $h$.
$m = \lim\limits_{h \to 0} -h^{2} - 3h -2$
Replace $h$ for $0$.
$m = \lim\limits_{h \to 0} -0^{2} - 3(0) -2 = -2$
b. Find the tangent line
Given: $m = -2$, $x = 1$, $y = 0$
Formula: $y = mx + b$
$0 = -2(1) + b$
$b = 2$
Tangent line: $y = -2x + 2$
c. This graph is the normal graph of the function and the tangent line.
