Question

Verify Lagrange's mean value theorem for the following function on the indicated interval. In each case find a point ′c′ in the indicated interval as stated by the Lagrange's mean value theorem:
f(x) = (x−1) (x−2) (x−3) on [0,4]

Answer 1

Hint: To solve the question, we have to apply Lagrange's mean value theorem condition for a function. We have to simplify the given function for calculating its derivative which is required while verifying the theorem.

Complete step-by-step answer:
The given function is f(x) = (x−1) (x−2) (x−3)

& f\left( x \right)=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)~ \\\ & =\left( x-1 \right)\left( {{x}^{2}}-2x-3x+6 \right) \\\ & =\left( x-1 \right)\left( {{x}^{2}}-5x+6 \right) \\\ & ={{x}^{3}}-5{{x}^{2}}+6x-{{x}^{2}}+5x-6 \\\ & ={{x}^{3}}-6{{x}^{2}}+11x-6 \\\ \end{aligned}$$ We know that a function f(x) on interval (a, b) such that b > a, is verified by Lagrange's mean value theorem when $${{f}^{1}}(c)=\dfrac{f(b)-f(a)}{b-a}$$ such that c lies in the interval of [a, b]. …. (1) We know that $${{f}^{1}}(x)=\dfrac{df(x)}{dx},\dfrac{d\left( f(x)+g(x) \right)}{dx}=\dfrac{df(x)}{dx}+\dfrac{dg(x)}{dx}$$. Thus, we get $$\begin{aligned} & {{f}^{1}}(x)=\dfrac{d\left( {{x}^{3}}-6{{x}^{2}}+11x-6 \right)~}{dx} \\\ & =\dfrac{d\left( {{x}^{3}} \right)}{dx}+\dfrac{d\left( -6{{x}^{2}} \right)}{dx}+\dfrac{d\left( 11x \right)}{dx}+\dfrac{d\left( -6 \right)}{dx} \\\ \end{aligned}$$ We know that $$\dfrac{d\left( cf(x) \right)}{dx}=c\dfrac{df(x)}{dx},\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$$ and differentiation of a constant is equal to 0. Thus, we get $$\begin{aligned} & =3{{x}^{3-1}}-6\dfrac{d\left( {{x}^{2}} \right)}{dx}+11\dfrac{d\left( x \right)}{dx}+0 \\\ & =3{{x}^{2}}-6\left( 2{{x}^{2-1}} \right)+11\left( 1 \right) \\\ & =3{{x}^{2}}-6\left( 2x \right)+11 \\\ & =3{{x}^{2}}-12x+11 \\\ \end{aligned}$$ Thus, we get the value of $${{f}^{1}}(c)=3{{c}^{2}}-12c+11$$ …. (2) The value of f(0) and f(4) by substituting x = 0 and x = 4 in f(x) = (x−1) (x−2) (x−3) $$\begin{aligned} & f(0)=\left( 0-1 \right)\left( 0-2 \right)\left( 0-3 \right) \\\ & =\left( -1 \right)\left( -2 \right)\left( -3 \right) \\\ & =2(-3) \\\ & =-6 \\\ \end{aligned}$$ Thus, the value of f(0) = -6 $$\begin{aligned} & f(4)=\left( 4-1 \right)\left( 4-2 \right)\left( 4-3 \right) \\\ & =\left( 3 \right)\left( 2 \right)\left( 1 \right) \\\ & =2(3) \\\ & =6 \\\ \end{aligned}$$ Thus, the value of f(4) = 6 We know that the value of $$\dfrac{f(b)-f(a)}{b-a}=\dfrac{f(4)-f(0)}{4-0}$$ Since for the given equation b = 4, a = 0. Thus, we get $$\dfrac{f(b)-f(a)}{b-a}=\dfrac{6-(-6)}{4-0}=\dfrac{6+6}{4}=\dfrac{12}{4}=3$$ Thus, we get the value of $$\dfrac{f(b)-f(a)}{b-a}$$ is equal to 3. From the equations (1) and (2), we get $$\begin{aligned} & 3{{c}^{2}}-12c+11=3 \\\ & 3{{c}^{2}}-12c+11-3=0 \\\ & 3{{c}^{2}}-12c+8=0 \\\ \end{aligned}$$ We know that the solution for $$a{{x}^{2}}+bx+c=0$$ is given by $$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$. Thus, the by applying the above formula for the quadratic equation in c, we get $$\begin{aligned} & c=\dfrac{-(-12)\pm \sqrt{{{(-12)}^{2}}-4(3)(8)}}{2(3)} \\\ & =\dfrac{12\pm \sqrt{144-96}}{6} \\\ & =\dfrac{12\pm \sqrt{48}}{6} \\\ & =\dfrac{12\pm \sqrt{16\times 3}}{6} \\\ & =\dfrac{12\pm 4\sqrt{3}}{6} \\\ & =\dfrac{6\pm 2\sqrt{3}}{3}\in (0,4) \\\ & \therefore c\in (0,4) \\\ \end{aligned}$$ Thus, Lagrange's mean value theorem is verified. Note: The possibility of mistake can be, not applying the differentiation rules for derivation of the given function. The other possibility of mistake can be, not applying the formula for calculating the roots of the quadratic equation, which is required for verifying the Lagrange's mean value theorem.