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Mathematics Question on Continuity and differentiability

Verify Mean Value Theorem,if f(x)=x3−5x2−3xf(x)=x^3-5x^2-3x in the interval [a,b][a,b],where a=1a=1 and b=3b=3.Find all c∈(1,3)c∈(1,3) for which f′(c)=0f'(c)=0

Answer

The given function is f(x)=x3−5x2−3xf(x)=x^3-5x^2-3x
ff, being a polynomial function, is continuous in [1,3][1,3] and is differentiable in (1,3)(1,3) whose derivative is 3x2−10x−3.3x^2−10x−3.
f(1)=13−5(1)2−3×1=−7f(1)=1^3-5(1)^2-3\times1=-7
f(3)=(3)3−5×(3)2−3×3=−27f(3)=(3)^3-5\times(3)^2-3\times3=-27
∴f(b)−f(a)b−a=f(3)−f(1)3−1∴\frac{f(b)-f(a)}{b-a}=\frac{f(3)-f(1)}{3-1}
=−27−(−7)3−1=−10=\frac{-27-(-7)}{3-1}=-10
Mean Value Theorem states that there is a point c∈(1,3)c∈(1,3) such that f′(c)=−10f'(c)=-10
f′(c)=−10f'(c)=-10
⇒3c2−10c−3=10⇒3c^2-10c-3=10
⇒3c2−10c+7=0⇒3c^2-10c+7=0
⇒3c2−3c−7c+7=0⇒3c^2-3c-7c+7=0
⇒3c(c−1)−7(c−1)=0⇒3c(c-1)-7(c-1)=0
⇒(c−1)(3c−7)=0⇒(c-1)(3c-7)=0
⇒c=1,73⇒c=1,\frac{7}{3},where c=73∈(1,3)c=\frac{7}{3}∈(1,3)
Hence, Mean Value Theorem is verified for the given function and c=73∈(1,3)c=\frac{7}{3}∈(1,3) is the only point for which f′(c)=0f'(c)=0