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Difference Quotients

Sol 1

\begin{displaymath}\frac{f(x+h)-f(x)}{h}=\frac{[2(x+h)^2-5(x+h)-9]-[2x^2-5x-9]}{h}\end{displaymath}


\begin{displaymath}=\frac{[2(x^2+2hx+h^2)-5(x+h)-9]-[2x^2-5x-9]}{h}\end{displaymath}


\begin{displaymath}=\frac{2x^2+4hx+2h^2-5x-5h-9-2x^2+5x+9}{h}=\frac{4hx+2h^2-5h}{h}\end{displaymath}


\begin{displaymath}=\frac{h(4x+2h-5}{h}=4x+2h-5\end{displaymath}

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Sol 2

\begin{displaymath}\frac{f(x+h)-f(x)}{h}=\frac{(x+h)^3-x^3}{h}\end{displaymath}


\begin{displaymath}=\frac{(x^3+3x^2h+3xh^2+h^3)- x^3}{h}=\frac{3x^2h+3xh^2+h^3}{h}\end{displaymath}


\begin{displaymath}=\frac{h(3x^2+3xh+h^2)}{h}=3x^2+3xh+h^2\end{displaymath}

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Sol 3

\begin{displaymath}\frac{f(x+h)-f(x)}{h}=\frac{\frac{1}{x+h}-\frac{1}{x}}{h}\end{displaymath}


\begin{displaymath}=\frac{x-(x+h)}{hx(x+h)}\end{displaymath}

(multiplying by $x(x+h)$ on the top and bottom)


\begin{displaymath}=\frac{-h}{hx(x+h)}=\frac{-1}{x(x+h)}\end{displaymath}

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Sol 4

\begin{displaymath}\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{x+h}-\sqrt{x}}{h}\end{displaymath}


\begin{displaymath}=\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}\end{displaymath}

(multiplying by $\sqrt{x+h}+\sqrt{x}$ on the top and bottom)


\begin{displaymath}=\frac{h}{h(\sqrt{x+h}+\sqrt{x})}=\frac{1}{\sqrt{x+h}+\sqrt{x}}\end{displaymath}

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Sol 5

\begin{displaymath}\frac{f(t)-f(x)}{t-x}=\frac{\frac{1}{t^2}-\frac{1}{x^2}}{t-x}\end{displaymath}


\begin{displaymath}\frac{x^2-t^2}{t^2x^2(t-x)}\end{displaymath}

(multiplying by $t^2x^2$ on the top and bottom)


\begin{displaymath}=\frac{(x-t)(x+t)}{t^2x^2(t-x)}=\frac{-(t-x)(x+t)}{t^2x^2(t-x)}\end{displaymath}


\begin{displaymath}= -\frac{x+t}{t^2x^2}\end{displaymath}

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Sol 6

\begin{displaymath}\frac{f(t)-f(x)}{t-x}=\frac{\frac{1}{2t+5}-\frac{1}{2x+5}}{t-x}\end{displaymath}


\begin{displaymath}\frac{(2x+5)-(2t+5)}{(t-x)(2t+5)(2x+5)}\end{displaymath}

(multiplying by $(2t+5)(2x+5)$ on the top and bottom)


\begin{displaymath}=\frac{2x-2t}{(t-x)(2t+5)(2x+5)}=\frac{-2(t-x)}{(t-x)(2t+5)(2x+5)}\end{displaymath}


\begin{displaymath}=\frac{-2}{(2t+5)(2x+5)}\end{displaymath}

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Sol 7

\begin{displaymath}\frac{f(t)-f(x)}{t-x}=\frac{\sqrt{4t^2+5}-\sqrt{4x^2+5}}{t-x}\end{displaymath}


\begin{displaymath}\frac{(4t^2+5)-(4x^2+5)}{(t-x)(\sqrt{4t^2+5}+\sqrt{4x^2+5})}\end{displaymath}

(multiplying by $(\sqrt{4t^2+5}+\sqrt{4x^2+5})$ on the top and bottom)


\begin{displaymath}=\frac{4t^2-4x^2}{(t-x)(\sqrt{4t^2+5}+\sqrt{4x^2+5})}\end{displaymath}


\begin{displaymath}=\frac{4(t-x)(t+x)} {(t-x)(\sqrt{4t^2+5}+\sqrt{4x^2+5})}\end{displaymath}


\begin{displaymath}=\frac{4(t+x)} {(\sqrt{4t^2+5}+\sqrt{4x^2+5})}\end{displaymath}

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Sol 8

\begin{displaymath}\frac{f(t)-f(x)}{t-x}=\frac{t^{2/3}-x^{2/3}}{t-x}\end{displaymath}


\begin{displaymath}=\frac{(t^{1/3})^2-(x^{1/3})^2}{(t^{1/3})^3-(x^{1/3})^3}\end{displaymath}


\begin{displaymath}=\frac{(t^{1/3}-x^{1/3})(t^{1/3}+x^{1/3})}{(t^{1/3}-x^{1/3})
((t^{1/3})^2+(t^{1/3})(x^{1/3})+(x^{1/3})^2))}\end{displaymath}


\begin{displaymath}=\frac{t^{1/3}+x^{1/3}}{t^{2/3}+t^{1/3}x^{1/3}+x^{2/3}}\end{displaymath}

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Lawrence Marx 2002-07-11