Question

1. 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 \(f(0)\):

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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 \(f^{\prime}(c)=3c^{2}+2 = 6\).

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Since \(c\in(0,2)\), we reject \(c =-\frac{2}{\sqrt{3}}\). So \(c=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\).

Answer:

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