Is #f(x) =(x+4)^3/(x^2/3+1)# concave or convex at #x=8#?

1 Answer
Mar 12, 2018

#f''(x) > 0# at #x=8#, so #f(x)# is concave upward at that point.

Explanation:

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