Let f(x) = (1/x^2) - (3/x) + 4. Find the intervals on which f is concave upward

Question

Let f(x) = (1/x^2) - (3/x) + 4. Find the intervals on which f is concave upward and those on which it is concave downward. Find all inflection points (b, f(b)) of f. Please show how to solve with details, don't just give the answer

Explanation / Answer

f(x) is concave upward when f '' (x) [double derivative] is > 0 f '' (x) = 6/x^4 - 6/x^3 f ''(x) > 0 ==> 6/x^4 - 6/x^3 > 0 ==> 6*(1-x)/x^4 > 0 ==> x 6/x^4 - 6/x^3 < 0 ==> 6*(1-x)/x^4 < 0 ==> x>1 infection point occurs at f ''(x) = 0 ==> 6/x^4 - 6/x^3 = 0 ==> 6*(1-x)/x^4 = 0 ==> x = 1 f(1) = 1-3+4 = 2 point of inflection is (1,2)

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