question 2\n5 pts\nfind the number c satisfying the conclusion of the mean value theorem for the function…

Explanation:

Step1: Recall Mean - Value Theorem formula

The Mean - Value Theorem states that if (y = f(x)) is continuous on the closed interval ([a,b]) and differentiable on the open interval ((a,b)), then (f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}), where (a=-3), (b = 0) and (f(x)=\frac{1}{3}x^{3}-x^{2}-3x + 5). First, find (f(a)) and (f(b)). [ \begin{align*} f(-3)&=\frac{1}{3}(-3)^{3}-(-3)^{2}-3(-3)+5\ &=\frac{1}{3}\times(-27)-9 + 9+5\ &=-9 - 9+9 + 5\ &=-4 \end{align*} ] [ \begin{align*} f(0)&=\frac{1}{3}(0)^{3}-(0)^{2}-3(0)+5\ &=5 \end{align*} ] Then (\frac{f(0)-f(-3)}{0-(-3)}=\frac{5-(-4)}{3}=\frac{9}{3}=3).

Step2: Find the derivative of (f(x))

Differentiate (f(x)=\frac{1}{3}x^{3}-x^{2}-3x + 5) using the power - rule ((x^{n})^\prime=nx^{n - 1}). [ \begin{align*} f^{\prime}(x)&=\frac{d}{dx}(\frac{1}{3}x^{3}-x^{2}-3x + 5)\ &=\frac{1}{3}\times3x^{2}-2x-3\ &=x^{2}-2x - 3 \end{align*} ]

Step3: Set (f^{\prime}(c)) equal to (\frac{f(0)-f(-3)}{0 - (-3)}) and solve for (c)

Set (x^{2}-2x - 3=3), which can be rewritten as (x^{2}-2x-6 = 0). Using the quadratic formula (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for the quadratic equation (ax^{2}+bx + c = 0) (here (a = 1), (b=-2), (c=-6)). [ \begin{align*} c&=\frac{2\pm\sqrt{(-2)^{2}-4\times1\times(-6)}}{2\times1}\ &=\frac{2\pm\sqrt{4 + 24}}{2}\ &=\frac{2\pm\sqrt{28}}{2}\ &=\frac{2\pm2\sqrt{7}}{2}\ &=1\pm\sqrt{7} \end{align*} ] Since (c\in[-3,0]), we take (c = 1-\sqrt{7}).

Answer:

(c = 1-\sqrt{7})