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# Pls help :)

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Let $f$ be defined by $f(x) = \left\{ \begin{array}{cl} 3-x & \text{ if } x \leq 3, \\ -x^3+2x^2+3x & \text{ if } x>3. \end{array} \right.$Calculate $f^{-1}(0)+f^{-1}(6)$.

My go at it:

First I attacked the $f^{-1}(0)$, I found the inverse which turned out to be exactly the same as the non inverse function so because 0 is less than 3, 3-0=3,

so it is $0+f^{-1}(6)$

$f^{-1}(6)$ is the $-x^3+2x^2+3x$, maybe it's a coincident but it factors into $-x(x+1)(x-3)$

Jan 24, 2021