Question
It is given that the Rolle's theorem holds for the function f(x) = x3 + bx2 + cx, $\text{x}\in[1,2]$ at the point $\text{x}=\frac{4}{3},$ the values of b and c.

Answer

So, f(1) = f(2)
⇒ (1)3 + b(1)2 + c(1) = (2)3 + b(2)2 + c(2)
⇒ 1 + b + c = 8 + 4b + 2c
⇒ 3b + c + 7 = 0 ....(i)
And $\text{f}'\Big(\frac{4}{3}\Big)=0$
$\Rightarrow3\Big(\frac{4}{3}\Big)^2+2\text{b}\Big(\frac{4}{3}\Big)+\text{c}=0$ [As, f'(x) = 3x2 + 2bx + c]
$\Rightarrow\frac{16}{3}+\frac{8\text{b}}{3}+\text{c}=0$
$\Rightarrow8\text{b}+3\text{c}+16=0\ ....(\text{ii})$
(ii) - (i) × 3, we get
8b - 9b + 16 - 21 = 0
⇒ -b - 5 = 0
⇒ b = -5
Substituting b = -5 in (i), we get
3(-5) + c + 7 = 0
⇒ -15 + c + 7 = 0
⇒ c = 8

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