Let f be defined by f(x) = x^2*ln(x).?
Find a number b so that f is concave in one direction on each of the intervals (0,b) and (b,infinity). Determine if f is concave up or down on each interval.
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What you're looking for is called an inflection point. To find an inflection point we have to find the second derivative of our original function and set it equal to 0 to solve for x.
f(x) = x^2 * ln(x)
f '(x) = 2x*ln(x) + x
f ''(x) = 2ln(x) + 3
Setting it equal to 0, we get:
2ln(x) + 3 = 0
2ln(x) = -3
ln(x) = -3/2
x = e^(-3/2)
x = 0.22313
Therefore, on the interval [ 0 , e^(-3/2) ], f(x) is concave down.
On the interval [ e*(-3/2) , infinity ], f(x) is concave up.
Hope this helped.