Question

If the function $$f(x) = x^4 - 2x^3 + ax^2 + bx$$ on $$[1,3]$$ satisfies the conditions of Rolle's Theorem with $$c = \dfrac{3}{2}$$, then find $$a$$ and $$b$$.

Solution

Given $$f(x)=x^4-2x^3+ax^2+bx$$ on $$[1,3]$$$$\implies f'(x)=4x^3-6x^2+2ax+b$$According to Rolle's theorem, if f(x) is continuous on $$[a,b]$$, differentiable on $$(a,b)$$ and $$f(a)=f(b)$$ then there exists some $$c \in (a,b)$$ such that $$f'(c)=0$$Given that the function $$f(x)$$ satisfies the conditions of Rolle's theorem with $$c=\dfrac 32$$$$f(1)=1^4-2(1)^3+a(1)^2+b(1)=1-2+a+b=a+b-1$$$$f(3)=3^4-2(3)^3+a(3)^2+b(3)=81-54+9a+3b=9a+3b-27$$from Rolles's theorem, $$f(1)=f(3)$$$$\implies a+b-1=9a+3b-27$$$$\implies 8a+2b=26$$$$\implies 4a+b=13$$ .......(1)from Rolles's theorem, $$f'(c)=4c^3-6c^2+2ac+b=0$$ $$\implies f'(\dfrac 32)=4(\dfrac 32)^3-6(\dfrac 32)^2+2a(\dfrac 32)+b=0$$$$\implies \dfrac{27}{2}-\dfrac{27}{2}+3a+b=0$$$$\implies 3a+b=0$$ .......(2)(1) - (2) $$\implies a=13$$substituting $$a=13$$ in (1) $$\implies 4(13)+b=13$$$$\implies b=-39$$Therefore, $$a=13,b=-39$$Mathematics

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