Answer
The equation of the tangent line is $y=-\frac{3}{16}x-\frac{1}{2}$.
Work Step by Step
The slope $m$ of the tangent line of the curve $f(x)$ at $A(a,b)$ is also the slope of the curve at that point, which is calculated by $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$
$$y=f(x)=\frac{1}{x^3}\hspace{1cm}A\Big(-2,-\frac{1}{8}\Big)$$
1) Find the slope $m$ of the tangent:
$$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$
Here $a=-2$ and $f(a)=b=-\frac{1}{8}$
$$m=\lim_{h\to0}\frac{f(h-2)-(-\frac{1}{8})}{h}=\lim_{h\to0}\frac{\frac{1}{(h-2)^3}+\frac{1}{8}}{h}$$ $$m=\lim_{h\to0}\frac{\frac{8+(h-2)^3}{8(h-2)^3}}{h}=\lim_{h\to0}\frac{8+(h^3-6h^2+12h-8)}{8h(h-2)^3}$$ $$m=\lim_{h\to0}\frac{h^3-6h^2+12h}{8h(h-2)^3}=\lim_{h\to0}\frac{h^2-6h+12}{8(h-2)^3}$$ $$m=\frac{0^2-6\times0+12}{8\times(0-2)^3}=\frac{12}{8\times(-8)}=\frac{12}{-64}=-\frac{3}{16}$$
2) Find the equation of the tangent line at $A(−2,−1/8)$:
The tangent line would have this form: $$y=-\frac{3}{16}x+m$$
Substitute $A(−2,−1/8)$ here to find $m$:
$$-\frac{3}{16}\times(-2)+m=-\frac{1}{8}$$ $$\frac{3}{8}+m=-\frac{1}{8}$$ $$m=-\frac{1}{2}$$
So the equation of the tangent line is $y=-\frac{3}{16}x-\frac{1}{2}$.
