Question

Let g be a twice-differentiable function with g(1) = 7 and g(7) = 1 and let f(x) = g(g(x)). a) Justify why there must be a value c, where 1 < c < 7, such that g' (c) = −1. b) Justify that f′ (1) = f′(7) and that there exists a value n such that on (1,7) where f''(n) = 0

Answer 1

Item a):

The Mean Value Theorem states that if a function f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point c in the interval (a,b) such that f'(c) is equal to the function's average rate of change over [a,b].

`f^(\prime)(c)=(f(b)-f(a))/(b-a)`

Applying this theorem in our problem, we have:

`g^(\prime)(c)=(g(7)-g(1))/(7-1)=(1-7)/(6)=-(6)/(6)=-1`

item b):

f(x) is given by the composition of g(x) with itself, therefore, we can calculate the derivative of f(x) by using the chain rule:

`f^(\prime)(x)=g^(\prime)(g(x))\cdot g^(\prime)(x)`

Then, evaluating this expression at x = 1 and x = 7, we have:

`\begin{gathered} f^(\prime)(1)=g^(\prime)(g(1))\cdot g^(\prime)(1)=g^(\prime)(7)\cdot g^(\prime)(1) \\ f^(\prime)(7)=g^(\prime)(g(7))\cdot g^(\prime)(7)=g^(\prime)(1)\cdot g^(\prime)(7) \end{gathered}`

Thus

`f^(\prime)(1)=f^(\prime)(7)`

Using the Leibniz rule, we can find the second derivative of f(x):

`f^(\prime)^(\prime)(x)=g^(\prime)^(\prime)(g(x))(g^(\prime)(x))^2+g^(\prime)(g(x))g^(\prime)^(\prime)(x)`

When those two terms are antisymmetric, the second derivative is equal to zero.