## Thomas' Calculus 13th Edition

$y=2x+3$, as shown in the figure.
Given the curve as $y=1/x^2$ and a point $(-1,1)$ on the curve, we can find the slope of the tangent line at this point as: $m=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}=\lim_{h\to0}\frac{1/(-1+h)^2-1/(-1)^2}{h}=\lim_{h\to0}\frac{1-(h-1)^2}{h(h-1)^2}=\lim_{h\to0}\frac{2h-h^2}{h(h-1)^2}=\lim_{h\to0}\frac{2-h}{(h-1)^2}=\frac{2-0}{(0-1)^2}=2$ The equation for the tangent line at $(-1,1)$ can then be written as $y-1=2(x+1)$ or $y=2x+3$ as shown in the figure.