Minimum value of $2^{x^{2} + 8 x + 6}$ occurs at x =

2

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- 2

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4

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- 4

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Solution

The correct option is D $- 4$
$f^{'} (x) = 2^{x^{2} + 8 x + 6} ((log 2) 2 x + 8)$
$= 2^{x^{2} + 8 x + 6} (log 2) (x + 4)^{2}$
$= 2^{x^{2} + 8 x + 7} (x + 4) {log}^{2}$
$f^{'} (x) = 0$ when $x = - 4$
$f^{'} (x) < 0$ $- \infty < x < - 4$
$f^{'} (x) > 0$ $x > - 4$
So $f (x)$ is minimum at $x = - 4$

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