Consider f: (0,2) => all real numbers defined by f(x) = x^x. Assume f has limit

Question

Consider f: (0,2) => all real numbers defined by f(x) = x^x. Assume f has limit at 0. Find it. (Hint: choose a sequence {Xn} converging to zero such that limit {f(Xn)} is easy to determine. Provide a detailed solution.

Explanation / Answer

a function is increasing only when it's derivative is strictly greater than 0. So take the derivative. f ' ( x ) = 6x ( x^2 - 1 )^2 set this greater than zero, and divide by 6 both sides. x ( x^2 - 1) ^2 > 0. Since the square is always greater than zero, then we know this function is increasing when x > 0. b) Relative min and max occur when the derivative is 0. f ' ( x ) = 6x ( x^2 - 1 )^2 = 0 so x = 0, and x = 1 , and x = -1 are the places when you have relative min's and max's. to find out which, plug in random values that are inbetween the min's and max's into the derivative to see whether your function is increasing or decreasing within that interval. First I'll pick a value which is less than -1, say -2. f ' ( -2 ) = -12 * ( 9 ) , this is a negative number, which means the slope is decreasing from ( -8 , -1) f ' ( - 1/2 ) = - 3 * ( 9 / 16 )this is a negative number, which means the slope is decreasing from ( -1, 0 ) f ' ( 1/2 ) is a positive number, which means the slope is increasing from ( 0 , 1 ) f ' ( 2 ) is a positive number, which means the slope is increasing from ( 1 , 8 ) From this analysis, looks like you have a relative minimum only at x = 0. Since the slope is decreasing before that point, and then increasing after that point. c ) concave upward is when the second derivative is > 0 starting with f ' ( x ) = 6x ( x^2 - 1 )^2 remember to do the product rule here. f '' ( x ) = 6 (x^2 - 1 )^2 + 24x^2 ( x^2 - 1 ) we want this > 0, but let's just find where the second derivative is equal to zero, and then plug in values into the intervals between to see whether it will be positive or negative in that interval. 6 (x^2 - 1 )^2 + 24x^2 ( x^2 - 1 ) = 0. divide everything by 6, and factor out an ( x^2 - 1 ) ( x^2 -1 ) [ x^2 -1 + 4 x^2 ] = 0 ( x^2 -1 )( 5x^2 - 1 ) = 0 So there are 4 places where f '' ( x ) = 0 x = 1 , x = -1 , x = 1 / v5 , x = - 1 / v5. v5 is a little greater than 2, so 1 / v5 is a little less than 1/2 we have 5 intervals to test. ( - 8 , -1 ) , ( -1 , - 1 / v5 ) , ( - 1 / v5 , 1 / v5 ) , ( 1 / v5 , 1 ) , ( 1 , 8 ) if I plug in -5 into the second derivative we get. f '' ( x ) = 6 ( x^2 -1 )( 5x^2 - 1 ) f '' ( -5 ) = 6 ( 24 )( 124 ) , this is a positive number, so your graph is concave up in this region. f '' ( - 1/2 ) = some negative number, so your graph is concave down in this region. f '' ( 0 ) = positive 6, so your graph is concave up in this region. f '' ( 1/2 ) = some negative number, so your graph is concave down in this region. f '' ( 10 ) some positive number, so your graph is concave up in this region. So the values for x in which your graph is concave up are, ( - 8 , -1 ) U ( - 1 / v5 , 1 / v5 ) U ( 1 , 8 )

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