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Question

It is given that the Rolle's theorem holds for the function $$f\left( x \right) ={ x }^{ 3 }+b{ x }^{ 2 }+cx,x\epsilon [1,2]$$ at the point $$x=\cfrac { 4 }{ 3 } $$. Find the values of $$b$$ and $$c$$.


Solution

Given $$f(x)=x^3+b x^2+c x$$

As we know that

Rolle's theorem states that if $$f(x)$$ be continuous on $$[a,b]$$,differentiable on $$(p,q)$$ and $$f(p)=f(q)$$ then there exists some $$r\in (p,q)$$ such that $$f'(r)=0$$

Given $$p=1,q=2$$ and $$r=\dfrac{4}{3}$$

$$f(1)=(1)^3+b(1)^2+c(1)=1+b+c$$

$$f(2)=(2)^3+b(2)^2+c(2)=8+4 b+2 c$$

According to Rolle's theorem 

$$f(1)=f(2)$$

$$\implies 1+b+c=8+4 b+2 c$$

$$\implies c=-7-3 b\cdots(1)$$

$$f'(x)=3 x^2+2 b x+c$$

According to Rolle's theorem 

$$f'(r)=f'\left(\cfrac43\right)=0$$

$$3 \bigg(\dfrac{4}{3}\bigg)^2+2 b\bigg(\dfrac{4}{3}\bigg)+c=0$$

$$\implies \dfrac{16}{3}+\dfrac{8 b}{3}+c=0$$

$$\implies 16+8 b+3 c=0$$

$$\implies 16+8 b+3(-7-3 b)=0 $$                $$(\because \text{from } (1))$$

$$\implies 16+8 b-21-9 b=0$$

$$\implies  b=-5$$

$$\implies c=-7-3 b=-7-3(-5)=15-7=8$$

Hence $$b=8,c=-5$$ 



Mathematics

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