14. find the value(s) c satisfying the mean value theorem for $f(x)=x^{3}+2x + 1$ on $0,2.$

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 = 0), (b = 2) for (y=f(x)=x^{3}+2x + 1). First, find (f(2)) and (f(0)): [ \begin{align*} f(2)&=2^{3}+2\times2 + 1\ &=8 + 4+1\ &=13 \end{align*} ] [ \begin{align*} f(0)&=0^{3}+2\times0 + 1\ &=1 \end{align*} ] Then (\frac{f(2)-f(0)}{2 - 0}=\frac{13 - 1}{2}=\frac{12}{2}=6).

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

Differentiate (f(x)=x^{3}+2x + 1) with respect to (x). Using the power - rule ((x^{n})^\prime=nx^{n - 1}), we have (f^{\prime}(x)=3x^{2}+2).

Step3: Set (f^{\prime}(c)) equal to (\frac{f(2)-f(0)}{2 - 0}) and solve for (c)

Set (f^{\prime}(c)=3c^{2}+2 = 6). [ \begin{align*} 3c^{2}+2&=6\ 3c^{2}&=4\ c^{2}&=\frac{4}{3}\ c&=\pm\frac{2}{\sqrt{3}} \end{align*} ] Since (c\in(0,2)), we reject (c =-\frac{2}{\sqrt{3}}). So (c=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}).

Answer:

(c=\frac{2\sqrt{3}}{3})