If the function $f (x) = 3 x^{4} - 8 x^{3} + 12 x^{2} - 48 x + 25$ has global maximum value $M$ and global minimum value $m$ in $[0, 3],$ then $2 M + m$ is equal to

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Solution

$f (x) = 3 x^{4} - 8 x^{3} + 12 x^{2} - 48 x + 25$
$\Rightarrow f^{'} (x) = 12 x^{3} - 24 x^{2} + 24 x - 48 = 12 (x^{3} - 2 x^{2} + 2 x - 4) = 12 (x - 2) (x^{2} + 2)$
$f^{'} (x) = 0$
$\Rightarrow x = 2 \in [0, 3]$
$f (0) = 25 \to M$
$f (2) = - 39 \to m$
$f (3) = 16$
So, $2 M + m = 50 - 39 = 11$

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