6.8 Practice Problems
Problem 6.1. The functions \(f\) and \(g\) are defined by \[f(x)=\frac {1}{x+2},\ x\neq -2,\qquad g(x)=\frac {x}{x-1},\ x\neq 1 .\] Find \(\left (g\circ f\right )^{-1}(x)\) and state its domain.
Show solution
Solution. Compose first. Applying \(f\) then \(g\), \[\left (g\circ f\right )(x)=g\left (\frac {1}{x+2}\right ) =\frac {\frac {1}{x+2}}{\frac {1}{x+2}-1} =\frac {1}{1-(x+2)}=\frac {-1}{x+1},\] multiplying above and below by \(x+2\) to clear the compound fraction.
Now invert. Put \(y=\frac {-1}{x+1}\) and solve for \(x\): \[y(x+1)=-1 \implies x+1=-\frac {1}{y} \implies x=-\frac {1}{y}-1 .\] Hence \[\left (g\circ f\right )^{-1}(x)=-\frac {1}{x}-1=-\frac {1+x}{x}, \quad x\neq 0 .\]
The domain of the inverse is the range of \(g\circ f\), and \(\frac {-1}{x+1}\) takes every real value except \(0\) — a fraction with constant numerator is never zero. So the domain is \(\mathbb {R}\setminus \{0\}\), agreeing with what the formula itself requires.
Note 6.30. Two checks are available and both are quick. Numerically, \(\left (g\circ f\right )(1)=-\frac {1}{2}\), and substituting \(-\frac {1}{2}\) into the inverse gives \(2-1=1\), returning the input.
Structurally, the domain of \(f^{-1}\) is always the range of \(f\), and the range of \(f^{-1}\) the domain of \(f\) — inverting a function swaps the two. That gives an independent way to find the domain of an inverse when the formula is awkward, and here the two methods agree.
- (a).
- When is a function said to be even? When is it odd?
- (b).
- Determine whether \(f(x)=6-7x^{2}\) is even, odd or neither.
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Solution. (a). \(f\) is even if \(f(-x)=f(x)\) for every \(x\) in its domain, and odd if \(f(-x)=-f(x)\).
(b). Replace \(x\) by \(-x\): \[f(-x)=6-7(-x)^{2}=6-7x^{2}=f(x),\] since \((-x)^{2}=x^{2}\). So \(f\) is even.
Note 6.31. “Neither” is the usual answer, not a failure. A function is even when its graph is symmetric about the \(y\)-axis and odd when it has rotational symmetry of half a turn about the origin; most functions have neither symmetry. \(f(x)=x^{2}+x\) is a quick example — \(f(-x)=x^{2}-x\), which is neither \(f(x)\) nor \(-f(x)\).
The one function that is both is \(f(x)=0\). And a useful shortcut for polynomials: one with only even powers (counting the constant as \(x^{0}\)) is even, one with only odd powers is odd, and one mixing them is neither. That settles \(6-7x^{2}\) at sight.
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