First, let's find $\left(f \circ f^{-1}\right)(x)$:
$\left(f \circ f^{-1}\right)(x) = f\left(f^{-1}(x)\right) = f\left(-\frac{1}{2}x\right)$
Now, we plug in the expression for $f(x)$:
$f\left(-\frac{1}{2}x\right) = -2\left(-\frac{1}{2}x\right) = x$
So, $\left(f
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