Question

video example 4
to illustrate the mean value theorem with a specific function, lets consider ( f(x)=x^{3}-x, a = 0, b = 4 ). since ( f ) is a polynomial, it is continuous and differentiable for all ( x ), so it is certainly continuous on ( 0,4 ) and differentiable on ( (0,4) ). therefore, by the mean value theorem, there is a number ( c ) in ( (0,4) ) such that
( f(4)-f(0)=f^{prime}(c)(4 - 0) ).
now ( f(4)=square, f(0)=square ), and ( f^{prime}(x)=square ), so this equation becomes
( square=f^{prime}(c)(4)=left(square
ight)(4)=square ),
which gives ( c^{2}=square ), that is, ( c=pmsquare ). but ( c ) must be in ( (0,4) ), so ( c=square ).
the following figure illustrates the calculation that the tangent line at this value of ( c ) is parallel to the secant line.

Explanation:

Step1: Calculate \(f(4)\) and \(f(0)\)

Given \(f(x)=x^{3}-x\), then \(f(4)=4^{3}-4 = 64 - 4=60\), \(f(0)=0^{3}-0 = 0\).

Step2: Find the derivative \(f^{\prime}(x)\)

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), for \(f(x)=x^{3}-x\), \(f^{\prime}(x)=3x^{2}-1\).

Step3: Substitute into the Mean - Value Theorem formula

The Mean - Value Theorem formula is \(f(4)-f(0)=f^{\prime}(c)(4 - 0)\). Substituting \(f(4) = 60\), \(f(0)=0\) and \(f^{\prime}(x)=3x^{2}-1\) into it, we get \(60=(3c^{2}-1)\times4\).

Step4: Solve the equation for \(c\)

First, divide both sides of \(60=(3c^{2}-1)\times4\) by \(4\): \(\frac{60}{4}=3c^{2}-1\), which simplifies to \(15 = 3c^{2}-1\). Then add \(1\) to both sides: \(15 + 1=3c^{2}\), so \(16 = 3c^{2}\), and \(c^{2}=\frac{16}{3}\). Then \(c=\pm\frac{4}{\sqrt{3}}=\pm\frac{4\sqrt{3}}{3}\). Since \(c\in(0,4)\), we take \(c = \frac{4\sqrt{3}}{3}\approx2.31\).

Answer:

\(f(4)=60\), \(f(0)=0\), \(f^{\prime}(x)=3x^{2}-1\), \(60=(3c^{2}-1)\times4\), \(c^{2}=\frac{16}{3}\), \(c=\pm\frac{4\sqrt{3}}{3}\), \(c=\frac{4\sqrt{3}}{3}\)