Question

Find maximum and minima value (if any) for:

 ∴f(x) = \(\begin{cases}max(g(t);t\underline{<}x)&;x<0\\ 3-x&;x>0\end{cases} \) where g(x) = x³ + x² + x + 1

Answer 1

g(x) = x³ + x² + x + 1

∴ g(t) = t³ + t² + t + 1

when x< 0 , f (x) = max (g(t) , t \(\underline {<}\) x)

∵ t \(\underline {<}\) x & x < 0

⇒ t < 0

now g¹ (t) = 3t² + 2t+1

= 3 (t² + 2/3 t + 1/3)

= 3((t + 1/3)² + 1/3 - 1/9)

= 3 (t + 1/3)² + 2/9)

= 3 (t + 1/3)² + 2/3 > 0

∴ g (t) is strictly increasing function

Hence, g (+) < g (0)   (∵ t < 0)

⇒ g (t) < 1   (∵ g (0) = 1)

∴ max g (t)  = 1 , t \(\underline {<}\) x & < 0

∴f(x) = \(\begin{cases}max(g(t);t\underline{<}x)&;x<0\\ 3-x&;x>0\end{cases} \)

\(\begin{cases}1&;x<0\\ 3-x&;x>0\end{cases} \)

∴ Maxima off(x) is 3.

& Minima does not exist