Question

Use the difference quotient and its limit h⇒0 to find the slope g(x)=8/x² at the point (2, 2); then find the equation of the tangent line there.

Answer 1

`g(x)=\cfrac{8}{x^2}\hspace{5em}\cfrac{dg}{dx}=\displaystyle \lim_(h\to 0)\cfrac{f(x+h)~~ - ~~f(x)}{h} \\\\[-0.35em] ~\dotfill\\\\ f(x+h)\implies \cfrac{8}{(x+h)^2}~~ - ~~\cfrac{8}{x^2}\implies \cfrac{8x^2~~ - ~~8(x+h)^2}{x^2(x+h)^2} \\\\\\ \cfrac{8x^2~~ - ~~8(x^2+2hx+h^2)}{x^2(x+h)^2}\implies \cfrac{8x^2-8x^2-16hx-8h^2}{x^2(x+h)^2} \\\\\\ \cfrac{-16hx-8h^2}{x^2(x+h)^2}\implies \cfrac{-8h(2x-h)}{x^2(x+h)^2} \\\\[-0.35em] ~\dotfill`

`\cfrac{f(x+h)~~ - ~~f(x)}{h}\implies \cfrac{ ~~ (-8h(2x-h))/(x^2(x+h)^2) ~~ }{h}\implies \cfrac{-8h(2x-h)}{hx^2(x+h)^2} \implies \cfrac{-8(2x-h)}{x^2(x+h)^2} \\\\[-0.35em] ~\dotfill\\\\ \displaystyle \lim_(h\to 0)\cfrac{f(x+h)~~ - ~~f(x)}{h}\implies \lim_(h\to 0) ~~ \cfrac{-8(2x-h)}{x^2(x+h)^2}\implies \lim_(h\to 0) ~~ \cfrac{-8(2x-0)}{x^2(x+0)^2} \\\\\\ \displaystyle \lim_(h\to 0) ~~\cfrac{-16x}{x^4}\implies -\cfrac{16}{x^3}`

now, we'd want to find the slope at (2,2) or namely when x = 2

`\left. -\cfrac{16}{x^3} \right|_(x=2)\implies -\cfrac{16}{2^3}\implies -2`

now for the tangent line at (2,2) we're really looking for the equation of a line whose slope is -2 and it passes through that point.

`(\stackrel{x_1}{2}~,~\stackrel{y_1}{2})\hspace{10em} \stackrel{slope}{m} ~=~ -2 \\\\\\ \begin{array}ll \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{-2}(x-\stackrel{x_1}{2}) \\\\\\ y-2=-2x+4\implies {\Large \begin{array}{llll} y=-2x+6 \end{array}}`