Question

Verify MVT (i.e., Mean Value Theorem) if $f\left(x\right)=x^2-4x-3$ in the interval [a, b], where a = 1 and b = 4.

Answer 1

Here, $f\left(x\right)=x^2-4x-3,xϵ[1,4]$ which is a polynomial function, so it is continuous and derivable at all $xϵR$,therefore

(i) f(x) is continuous on [1, 4] (ii) f(x) is derivable on (1, 4)

$\therefore$ Conditions of Lagrange's theorem are satisfied on [1, 4].

Hence, there is atleast one real number. C$ϵ$(1, 4) such that

$f^{\prime }\left(c\right)=\frac{f\left(4\right)-f\left(1\right)}{4-1}(\therefore f^{\prime }(c)=\frac{f\left(b\right)-f\left(a\right)}{b-a})\Rightarrow 2c-4=\frac{(4^2-4\times 4-3)-(1^2-4\times 1-3)}{4-1}=1(\because f^{\prime }(x)=\frac{d}{dx}(x^2-4x-3)=2x-4)\Rightarrow 2c-4=1\Rightarrow c=\frac{5}{2}ϵ(1,4)$