Question

Rolle’s theorem holds for the function $f\left( x \right)={{x}^{3}}+b{{x}^{2}}+cx$, $1\le x\le 2$ at the point $\left( x=\dfrac{4}{3} \right)$ then the value of b and c are

Answer 1

From the question given that we have been asked to find the value of b and c. as we know that Rolle’s theorem states that the if a function f is a continuous on the closed interval $\left[ a,b \right]$ and differentiable on the open interval $\left( a,b \right)$ such that $f\left( a \right)=f\left( b \right)$, then ${{f}’}\left( x \right)=0$ for some x with $a\le x\le b$. From this we will get the values of b and c.

Complete step-by-step solution:
From the question given that the Rolle’s theorem holds for the function,
$\Rightarrow f\left( x \right)={{x}^{3}}+b{{x}^{2}}+cx$
And x lies between,
$\Rightarrow 1\le x\le 2$
As we know that Rolle’s theorem states that the if a function f is a continuous on the closed interval $\left[ a,b \right]$ and differentiable on the open interval $\left( a,b \right)$ such that $f\left( a \right)=f\left( b \right)$, then ${{f}’}\left( x \right)=0$ for some x with $a\le x\le b$.
Here Rolle’s theorem holds on $x\in \left[ 1,2 \right]$at $x=\dfrac{4}{3}$ then we can say that,
From the Rolle’s theorem we can say that,
$\Rightarrow f\left( 1 \right)=f\left( 2 \right)$
Now by substituting the values in the function we will get,
$\Rightarrow 1+b+c=8+4b+2c$
Now shifting the all the left-hand side terms to the right-hand side, we will get,
$\Rightarrow 8+4b+2c-1-b-c=0$
Now by further simplification we will get,
$\Rightarrow 7+3b+c=0…………………………..\left( 1 \right)$
Now again from Rolle’s theorem we will get that,
$\Rightarrow {{f}’}\left( \dfrac{4}{3} \right)=0$
Now by differentiating the function we will get,
$\Rightarrow {{f}’}\left( x \right)=3{{x}^{2}}+2bx+c=0$
Now ${{f}’}\left( \dfrac{4}{3} \right)=0$ by substituting we will get,
$\Rightarrow {{f}’}\left( \dfrac{4}{3} \right)=3{{\left( \dfrac{4}{3} \right)}^{2}}+2b\left( \dfrac{4}{3} \right)+c=0$
Now by further simplifying we will get,
$\Rightarrow 8b+3c+16=0…………………..\left( 2 \right)$
Now we have to solve the equation $\left( 1 \right)$and $\left( 2 \right)$
Multiplying the equation $\left( 1 \right)$ with $3$ we will get,
$\Rightarrow 9b+3c+21=0$
and then subtract the both the equation $\left( 1 \right)$and $\left( 2 \right)$
$\begin{aligned} & \Rightarrow 9b+3c+21=0 \\\ & \left( - \right)\,8b+3c+16=0 \\\ \end{aligned}$
$\Rightarrow b+5=0$
$\Rightarrow b=-5$
Now by substituting the value of b in equation $\left( 1 \right)$we will get the value of c,
$\Rightarrow 3b+c+7=0$
$\Rightarrow 3\left( -5 \right)+c+7=0$
Now by further simplification we will get,
$\Rightarrow c-8=0$
$\Rightarrow c=8$
Therefore, the values of b and c are $b=-5$ and $c=8$.

Note: Students should know the Rolle’s theorem conditions and students should be very careful if students did not do differentiation of function and substituted the value of x and equated to zero the whole answer will be wrong, that means students did $f\left( \dfrac{4}{3} \right)=0$ instead of ${{f}^{|}}\left( \dfrac{4}{3} \right)=0$ the whole answer will be wrong.