Calculus/Precalculus/Solutions

Current revision (unreviewed)

[edit] Functions

1. $f (0) = 0$ , $f (2) = 4$
2. The domain is $(-\infty,\infty)$ ; the range is $[0,\infty)$ ,
3. No, since $f$ isn't one-to-one; for example, $f ( - 1) = f (1) = 1$ .
1. 1. $(f + g)(x) = x + 2 + 1 / x = (x 2 + 2 x + 1) / x$ .
  2. $(f - g)(x) = x + 2 - 1 / x = (x 2 + 2 x - 1) / x$ .
  3. $(g - f)(x) = 1 / x - x - 2 = (1 - x 2 - 2 x) / x$ .
  4. $(f\times g)(x)=(x+2)/x$ .
  5. $(f / g)(x) = x (x + 2)$ provided $x\ne0$ . Note that 0 is not in the domain of $f / g$ , since it's not in the domain of $g$ , and you can't divide by something that doesn't exist!
  6. $(g / f)(x) = 1 / [x (x + 2)]$ . Although 0 is still not in the domain, we don't need to state it now, since 0 isn't in the domain of the expression $1 / [x (x + 2)]$ either.
  7. $(f\circ g)(x)=1/x+2=(2x+1)/x$ .
  8. $(g\circ f)(x)=1/(x+2)$ .
2. $f (g (2)) = 5 / 2$ ; $g (f (2)) = 1 / 4$ .
3. Yes; $f - 1 (x) = x - 2$ and $g - 1 (x) = 1 / x$ . Note that $g$ and its inverse are the same.
As pictured, by the Vertical Line test, this graph represents a function.