Answer to Question #94033 in Calculus for Jflows

Question

7.Find the number \\(c\\) guaranteed by the mean value theorem for derivatives for \\(f(x)=(x+1)^{3}, [-1, 1] \\)
a.\\(c=1\\pm \\sqrt(5)}\\)
b.\\(c=\\frac{-\\sqrt (3) \\pm 2}{\\sqrt(3)}\\)
c.\\(c=\\frac{-\\sqrt (5) \\pm 2}{\\sqrt(5)}\\)
d.\\(c=\\frac{-\\sqrt (2) \\pm 1}{\\sqrt(3)}\\)

8.Determine whether the mean value theorem can be applied to \\(f\\) on the closed interval [a, b] . If can be applied, Find the value of \\(c\\) in open interval (a, b) such that \\(f(x)=x(x^{2}-x-2), [-1, 1]\\)
a.\\(c=\\frac{-1}{3}\\)
b.\\(c=\\frac{-2}{3}\\)
c.\\(c=\\frac{-2}{5}\\)
d.\\(c=\\frac{-1}{2}\\)

Answer 1

7. According to Mean Value Theorem, there must exist $c$ s.t. $f'(c) = \frac{f(b) - f(a)}{b - a}$ . In our case, $f'(x) = 3 (x+1)^2$ , and $f(1) = 8, f(-1) = 0$ , hence one has to solve quadratic equation $3 (c+1)^2 = 4$ for $c$ . Obtain: $c = \frac{-3 \pm 2 \sqrt{3}}{3} = \frac{-\sqrt{3} \pm \sqrt{2}}{\sqrt{3}}$ , so the answer is b).

8 $f'(x) = x^2 - x - 2 + x(2 x - 1) = 3 x^2 - 2 x - 2$ . The function is obviously continuous and differentiable in $[-1,1]$, so we can apply Mean Value Theorem: $f'(c) = 3 c^2 - 2 c - 2 = \frac{f(1) - f(-1)}{2} = \frac{-2 - 0}{2} = -1$ . Solving for $c$ , obtain $c = -\frac{1}{3}, c = 1$, so the answer is a).