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Global Maxima

Minimum value...

Question

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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Similar questions

f (x) = 4 x^{4} + 12 x^{3} + c x^{2} + 6 x + d

- 4 \leq 8 x - 2 \leq 4

\frac{1}{x^{2}}

m_{1}

x^{2} - 6 x + 13

m_{2}

- x^{2} - 8 x + 4

m_{2} - m_{1} =

x^{2} + \frac{1}{1 + x^{2}}

x

\frac{x^{2} + x - 6}{x^{2} - 8 x + 12} = 4

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