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liraira [26]
3 years ago
8

F(x)=2x*e^2x. find lim of f(x) as x--> - infinity and +infinity ...?

Mathematics
1 answer:
attashe74 [19]3 years ago
4 0
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>
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