Question

example 3 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)=60 ), ( f(0)=0 ), and ( f^{prime}(x)=3 x^{2}-1 ), so this equation becomes ( 60=f^{prime}(c)(4)=left(3 c^{2}-1
ight) 4=12 c^{2}-4 ), which gives ( c^{2}=64 ), that is, ( c=pm \frac{16}{3} ). but ( c ) must be in ( (0,4) ), so ( c=\frac{4}{sqrt{3}} ). the figure illustrates this calculation: the tangent line at this value of ( c ) is parallel to the secant line.

Explanation:

Step1: Substitute into the equation

We have the equation \(60=(3c^{2}-1)\times4\). First, divide both sides of the equation by \(4\):
\(\frac{60}{4}=3c^{2}-1\)
\(15 = 3c^{2}-1\)

Step2: Solve for \(c^{2}\)

Add \(1\) to both sides of the equation:
\(15 + 1=3c^{2}\)
\(16=3c^{2}\)
Then divide both sides by \(3\):
\(c^{2}=\frac{16}{3}\)

Step3: Solve for \(c\)

Take the square root of both sides:
\(c=\pm\sqrt{\frac{16}{3}}=\pm\frac{4}{\sqrt{3}}\)
Since \(c\in(0,4)\), we take the positive value \(c = \frac{4}{\sqrt{3}}\)

Answer:

\(c^{2}=\frac{16}{3}\), \(c=\pm\frac{4}{\sqrt{3}}\) and the valid \(c\) in \((0,4)\) is \(c = \frac{4}{\sqrt{3}}\)