View source for Solutions To Mathematics Textbooks/Calculus Early Transcendentals (6th) (0495011665)/Chapter 1.1

=Chapter 1.1 Exercises=
==25==

Replace instances of <math>f(x)\,</math> with the expression <math>\frac{1}{x}</math>.


<math>
\frac{\frac{1}{x}-\frac{1}{a}}{x-a} = \left( \frac{1}{x}-\frac{1}{a} \right) \cdot \frac{1}{x-a}
</math>


<math>
= \frac{a-x}{xa(x-a)} = \frac{(a-x)}{-xa(a-x)} = -\frac{1}{ax}
</math>

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