It is given that the Rolles theorem holds for the function `f(x)=x^3+b x^2+c x ,x in [1,2]` at the point `x=4/3dot` Find the values of b and c dot

Answer

we get
`b=-5, c=8`

Explanation

If rolle's theorem holds on `x \in[1,2]` at `x=4 / 3` then we can say two things for sure,

`\rightarrow f(1)=f(2)`

`1+b+c=8+4 b+2 c`

`3 b+c+7=0 \quad \ldots \ldots . .(1)`


`\rightarrow f^{\prime}(4 / 3)=0`

`f^{\prime}(x)=3 x^{2}+2 b x+c=0` at `x=4 / 3`

`3(4 / 3)^{2}+2 b(4 / 3)+c=0`

`8 b+3 c+16=0 \quad \ldots \ldots . .(2)`

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