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

Chapter 1.1 Exercises

25

Replace instances of $f(x)\,$ with the expression ${\frac {1}{x}}$.

${\frac {{\frac {1}{x}}-{\frac {1}{a}}}{x-a}}=\left({\frac {1}{x}}-{\frac {1}{a}}\right)\cdot {\frac {1}{x-a}}$

$={\frac {a-x}{xa(x-a)}}={\frac {(a-x)}{-xa(a-x)}}=-{\frac {1}{ax}}$