Question

the function ( f ) is defined by ( f(x)=x^{3}+2 x - 3 cos (2 x) ). find all values of ( x ) that satisfy the conclusion of the mean value theorem on the interval ( -1,2 ). you may use a calculator and ro to the nearest thousandth.

Explanation:

Step1: Recall the Mean Value Theorem formula

The Mean Value Theorem states that if \(y = f(x)\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then \(f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}\). Here, \(a=-1\), \(b = 2\). First, find \(f(-1)\) and \(f(2)\).

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Then \(\frac{f(2)-f(-1)}{2-(-1)}=\frac{13.9608-(-1.7517)}{3}=\frac{15.7125}{3}=5.2375\)

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

Differentiate \(f(x)=x^{3}+2x - 3\cos(2x)\) using the sum rule and chain - rule. \(f^{\prime}(x)=3x^{2}+2+6\sin(2x)\)

Step3: Set \(f^{\prime}(x)\) equal to \(\frac{f(2)-f(-1)}{2 - (-1)}\)

Set \(3x^{2}+2+6\sin(2x)=5.2375\), so \(3x^{2}+6\sin(2x)-3.2375 = 0\)

Step4: Use a calculator to solve the equation

Using a graphing calculator or a numerical method (such as the Newton - Raphson method) to solve the equation \(y = 3x^{2}+6\sin(2x)-3.2375\) for \(x\in[-1,2]\).
Let \(g(x)=3x^{2}+6\sin(2x)-3.2375\), \(g^{\prime}(x)=6x + 12\cos(2x)\)
Starting with an initial guess (by graphing \(y = 3x^{2}+2+6\sin(2x)\) and \(y = 5.2375\)), and applying the Newton - Raphson formula \(x_{n + 1}=x_{n}-\frac{g(x_{n})}{g^{\prime}(x_{n})}\)
After several iterations (or using the calculator's equation - solving function), we find \(x\approx0.574\)

Answer:

\(x\approx0.574\)