Question

Verify Mean Value Theorem,if \(f(x)=x^3-5x^2-3x\) in the interval \([a,b]\),where \(a=1\) and \(b=3\).Find all \(c∈(1,3)\) for which \(f'(c)=0\)

Answer 1

The given function is \(f(x)=x^3-5x^2-3x\)
\(f\), being a polynomial function, is continuous in \([1,3]\) and is differentiable in \((1,3)\) whose derivative is \(3x^2−10x−3.\)
\(f(1)=1^3-5(1)^2-3\times1=-7\)
\(f(3)=(3)^3-5\times(3)^2-3\times3=-27\)
\(∴\frac{f(b)-f(a)}{b-a}=\frac{f(3)-f(1)}{3-1}\)
\(=\frac{-27-(-7)}{3-1}=-10\)
Mean Value Theorem states that there is a point \(c∈(1,3)\) such that \(f'(c)=-10\)
\(f'(c)=-10\)
\(⇒3c^2-10c-3=10\)
\(⇒3c^2-10c+7=0\)
\(⇒3c^2-3c-7c+7=0\)
\(⇒3c(c-1)-7(c-1)=0\)
\(⇒(c-1)(3c-7)=0\)
\(⇒c=1,\frac{7}{3}\),where \(c=\frac{7}{3}∈(1,3)\)
Hence, Mean Value Theorem is verified for the given function and \(c=\frac{7}{3}∈(1,3)\) is the only point for which \(f'(c)=0\)