I am assuming it is f(x)=2x e^2x <span>2x times e raised to the 2x power, and not 2 times e raised to the 2x** </span>
<span>as it approaches infinity, your f(x) --> infinity </span> <span>we can see by inspection, as 2x will approach inf, and e^(2x) will approach infinity even faster. </span> <span>inf x inf = inf </span>
<span>as it approaches neg infinity, your f(x) --> 0 </span> <span>we can see by inspection, as 2x will approach neg inf, and e^(2x) will approach zero. </span> <span>- inf x 0 = - 0 </span>
<span>The minimum(or maximum) is given when the derivative = 0 </span> <span>Using the product rule, </span> <span>f ' (x) = 2e^(2x) + 4x e(2x) </span> <span>f ' (x) = 2e^(2x) ( 1 + 2x ) </span> <span>We find the roots </span> <span>2e^(2x) will never equal 0 </span> <span>1 + 2x = 0, x = -1/2 </span>
<span>The minimum value will be when x = -0.5</span>