Question

Verify Lagrange’s mean value theorem for f(x)=x^3-x^2-5x+3 in [0, 4] and find the constant ‘c'​

Answer 1

$\tt f(x)=x^3-x^2-5x+3 \quad$ in [0,4] \). Being a polynomial f(x) is continuous in [0,4] and also derivable in (0,4). Here, $\tt f^{\prime}(x)=3 x^2 -2x-5$

⟹ Conditions of Lagrange's mean value theorem are satisfied for f(x) in [0,4].

$\color{olive} \begin{aligned} & \tt \Rightarrow \exists c \in(0,4) \\ \\ & \tt f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\ \\ & \tt=\frac{f(4)-f(0)}{4-0} \\ \\ \Rightarrow & \tt 3 c^{2}-2 c-5=\frac{\left [(4)^{3}-(4)^{2}-5(4)+3\right]}{4} \\ \\ \Rightarrow & \tt 3 c^{2}-2 c-5=\frac{6}{4}+20+3 \\ \\ \Rightarrow & \tt 3 c^{2}-2 c-5=\frac{31}{4} \\ \\ \Rightarrow & \tt 36^{2}-2 c-\frac{51}{4}=0 \\ \\ \Rightarrow & \tt C=\pm \frac{\sqrt{157}}{6}+\frac{1}{3} \: \: \: and \: \: \: \exists c \in(0,4)\end{aligned}$