Is #f(x) =(x-2)^3/(x^2-7)# concave or convex at #x=8#?

1 Answer
Psykolord1989 .
Oct 1, 2017

The function will be convex (aka concave upward) if #f''(8)>0#, and concave (concave downward) if #f''(8)<0#

We must use the quotient rule to find this derivative. The quotient rule states that given #f(x)=g(x)/(h(x)), f'(x) = (g'(x)h(x)-g(x)h'(x))/(h(x))^2#

#(df)/dx = (3(x-2)^2*(x^2-7) - 2x(x-2)^3)/((x^2-7)^2#

The product rule will help us find our next derivative, stating that #q(x) = r(x)s(x) -> (dq)/dx = r'(x)s(x) + r(x)s'(x)#

For our new setup, we have #g(x) = 3(x-2)^2*(x^2-7) - 2x(x-2)^3 -> g'(x) = 6(x-2)(x^2-7) + 6x(x-2)^2 - 2(x-2)^3 - 6x(x-2)^2 = 6(x-2)(x^2-7)-2(x-2)^3#

And #h(x) = (x^2-7)^2 -> h'(x) = 4x(x^2-7)#
#f''(x) = ((6(x-2)(x^2-7) -2(x-2)^3)(x^2-7)^2 - (4x(3(x-2)^2(x^2-7) -2x(x-2)^3)/(x^2-7)^4#

Related questions
Impact of this question
731 views around the world
You can reuse this answer
Creative Commons License