What is the extreme 4e^{2x^2+5x+16} ?

The extreme 4e^{2x^2+5x+16} is Minimum(-5/4 ,4e^{103/8})

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extreme 4e^{2x^2+5x+16}

extreme points

4 e^{2 x^{2} + 5 x + 16}

Minimum (- \frac{5}{4}, 4 e^{\frac{103}{8}})

Solution steps

f^{'} (x) = 4 e^{2 x^{2} + 5 x + 16} (4 x + 5)

Find intervals:

Decreasing

: - \infty < x < - \frac{5}{4},

Increasing

: - \frac{5}{4} < x < \infty

Plug

x = - \frac{5}{4}

into

4 e^{2 x^{2} + 5 x + 16} : 4 e^{\frac{103}{8}}

Minimum (- \frac{5}{4}, 4 e^{\frac{103}{8}})

f (x) = \frac{( x + 3 )}{( x - 2 )}

f (x) = x^{4} + 4 x^{2} - 10, - 5 \leq x \leq 2

f (x, y) = 3 x y^{2} + x^{3} - 3 x

\frac{x ^{2} - 9}{x ^{2} + 9}

f (x, y) = x^{2} + x y + y^{2} + x - y + 1

What is the extreme 4e^{2x^2+5x+16} ?
The extreme 4e^{2x^2+5x+16} is Minimum(-5/4 ,4e^{103/8})