Asked by Sam

x(x+4)^(2/3)

a). Find the second derivative
b). Find any points of Inflection
c). Determine the intervals of Concavity

Answers

Answered by Steve
y = x(x+4)^(2/3)

y' = (x+4)^(2/3) + 2x/3 * (x+4)^(-1/3)
= (5x+12)/(3(x+4)^(1/3))

y'' = 2(5x+24)/(9(x+4)^(4/3))

Since the denominator is always positive, we just need to look at the numerator.

inflection points where f '' = 0, x=-24/5

That's where it changes from concave down to concave up.

However, watch out for x = -4, where y' and y'' are undefined.
Answered by Reiny
1st deriv
= x(2/3)(x+4)^(-1/3) + (x+4)^(2/3)
= (1/3)(x+4)^(-1/3) [ 2x+ 3(x+4) ]
= (1/3)(5x+12)(x+4)^(-1/3)

2nd deriv
= (1/3)(5x+12)(-1/3)(x+4)^(-4/3) + (1/3)(5))x+4)^(-1/3)
= ....
= 2(5x+24)/(9(x+4)^(4/3) )

should be easy from there
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