Respuesta :
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].
[tex]f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}[/tex]Applying this theorem in our problem, we have:
[tex]g^{\prime}(c)=\frac{g(7)-g(1)}{7-1}=\frac{1-7}{6}=-\frac{6}{6}=-1[/tex]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:
[tex]f^{\prime}(x)=g^{\prime}(g(x))\cdot g^{\prime}(x)[/tex]Then, evaluating this expression at x = 1 and x = 7, we have:
[tex]\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}[/tex]Thus
[tex]f^{\prime}(1)=f^{\prime}(7)[/tex]Using the Leibniz rule, we can find the second derivative of f(x):
[tex]f^{\prime}^{\prime}(x)=g^{\prime}^{\prime}(g(x))(g^{\prime}(x))^2+g^{\prime}(g(x))g^{\prime}^{\prime}(x)[/tex]When those two terms are antisymmetric, the second derivative is equal to zero.
