Question

the function ( f ) is defined by ( f(x)=x^{2}+2 - 5sin(x^{2}) ). find all values of ( x ) that satisfy the conclusion of the mean value theorem on the interval (0,2). you may use a calculator and round to the nearest thousandth.

Explanation:

Step1: Check the conditions of the Mean Value Theorem

The function \(f(x)=x^{2}+2 - 5\sin(x^{2})\) is continuous on \([0,2]\) (since \(y = x^{2}\), \(y=\sin t\) (where \(t=x^{2}\)) are continuous functions and sums and scalar - multiples of continuous functions are continuous) and differentiable on \((0,2)\) (since \(y = x^{2}\), \(y=\sin t\) (where \(t = x^{2}\)) are differentiable functions and sums and scalar - multiples of differentiable functions are differentiable).

Step2: Calculate \(f(0)\) and \(f(2)\)

• \(f(0)=0^{2}+2-5\sin(0^{2})=2\)
• \(f(2)=2^{2}+2-5\sin(2^{2})=4 + 2-5\sin(4)=6-5\sin(4)\)

Step3: Calculate the derivative \(f^{\prime}(x)\)

Using the chain rule, if \(y=x^{2}+2-5\sin(x^{2})\), then \(f^{\prime}(x)=2x-10x\cos(x^{2})\)

Step4: Apply the Mean Value Theorem formula

The Mean Value Theorem states that \(f^{\prime}(c)=\frac{f(2)-f(0)}{2 - 0}\)

First, \(\frac{f(2)-f(0)}{2-0}=\frac{(6 - 5\sin(4))-2}{2}=\frac{4-5\sin(4)}{2}\)

We need to solve the equation \(2x-10x\cos(x^{2})=\frac{4 - 5\sin(4)}{2}\)

Let \(g(x)=2x-10x\cos(x^{2})-\frac{4 - 5\sin(4)}{2}\)

Using a calculator (for example, a graphing calculator or a numerical root - finding algorithm like the Newton - Raphson method):
\(\sin(4)\approx - 0.757\), then \(\frac{4-5\sin(4)}{2}=\frac{4+3.785}{2}=\frac{7.785}{2}=3.8925\)

The equation \(2x-10x\cos(x^{2})=3.8925\)

By using a calculator's root - finding function (on the interval \((0,2)\)):
\(x\approx1.676\)

Answer:

\(x = 1.676\)