One way to determine whether a function is convex or concave at a particular point is to evaluate its second derivative at that point.
Starting with:
f(x)=(x+4)^3/(x^2/3 + 1)
Let's visualize it as a product of two functions:
f(x)= u * v
u(x) = (x+4)^3
v(x) = (x^2/3 + 1)^(-1)
Applying the Chain Rule once:
f'(x)= uv' + vu'
and then applying it a second time, we have:
f''(x)= uv'' + v'u' + v'u' + v'u'' = uv'' + 2u'v' + u''v
We already have u and v above, so let's find the derivatives needed to form the three terms in f''(x):
u'(x) = 3(x+4)^2
u''(x) = 6(x+4)
v'(x) =(-1)(1/3)(2x)(x^2/3 + 1)^(-2) = -(2x)/3(x^2/3 + 1)^(-2)
For v''(x), we will need the Chain Rule one more time:
v'(x) = p * q
p(x) = -(2x)/3
q(x) = (x^2/3 + 1)^(-2)
p'(x) = -2/3
q'(x) = (-2)((2x)/3)(x^2/3 + 1)^(-3) = -(4x)/3(x^2/3 + 1)^(-3)
v''(x) = pq' + qp'
v''(x) = (8x^2)/9(x^2/3 + 1)^(-3) - 2/3(x^2/3 + 1)^(-2)
Now, we could substitute all of these pieces back into the equation for f''(x) but, since we are planning to evaluate this function at x=8, it is simpler (and less error-prone) to evaluate the pieces at this value and then combine the pieces into the whole:
u(8) = (8+4)^3 = 12^3 = 1728
u'(8) = 3(8+4)^2 = 3 * 12^2 = 432
u''(8) = 6(8+4) = 6 * 12 = 72
v(8) = (8^2/3 + 1)^(-1) = (64/3 + 3/3)^(-1) = (67/3)^(-1) = 3/67
v'(8) = -(2*8)/3(8^2/3 + 1)^(-2) = -16/3(64/3 + 3/3)^(-2)
=-16/3(67/3)^(-2) =-16/cancel(3) * (cancel(3)*3)/67^2 =-48/67^2
v''(x) = (8*8^2)/9(8^2/3 + 1)^(-3) - 2/3(8^2/3 + 1)^(-2)
=512/9(64/3 + 3/3)^(-3)-2/3(64/3 + 3/3)^(-2)
=512/cancel(3*3)*(3*cancel(3*3))/67^3 - 2/cancel(3)*(3*cancel(3))/67^2 = 1536/67^3 - (6 * 67)/67^3
= 1134/67^3
And finally:
f''(8)= u(8)v''(8) + 2u'(8)v'(8) + u''(8)v(8)
= 1728*(1134/67^3)
+2*432*(-48/67^2)(67/67)
+72*3/67(67^2/67^2)
=1959552/300763 - 2778624/300763 + 969624/300763
f''(8) = 150552/300763
Therefore, f''(x) > 0 at x=8, so f(x) is concave upward at that point.
SEE: http://mathsfirst.massey.ac.nz/Calculus/Sign2ndDer/Sign2DerPOI.htm
