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  • Verify mean value theorem for each of the functions given f(x) = x3 - 2x2 - x + 3 in [0, 1]

Verify mean value theorem for each of the functions given
f(x) = x^3 - 2x^2 - x + 3 in [0, 1]

Answers (1)

Given: f(x) = x^3 - 2x^2 - x + 3 in [0, 1]

Now, we have to show that f(x) verify the Mean Value Theorem

First of all, Conditions of Mean Value theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that

f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}

Condition 1:

f(x) = x^3 - 2x^2 - x + 3

Since, f(x) is a polynomial and we know that, every polynomial function is continuous for all x ∈ R

\Rightarrow f(x) = x^3 - 2x^2 - x + 3  is continuous at x ∈ [0,1]

Hence, condition 1 is satisfied.

Condition 2:

\Rightarrow f(x) = x^3 - 2x^2 - x + 3

Since, f(x) is a polynomial and every polynomial function is differentiable for all x ∈ R

f^{\prime}(x)=3 x^{2}-4 x-1

⇒ f(x) is differentiable at [0,1]

Hence, condition 2 is satisfied.

Thus, Mean Value Theorem is applicable to the given function

Now,

\begin{aligned} &f(x)=x^{3}-2 x^{3}-x+3 x \in[0,1]\\ &f(a)=f(0)=3\\ &f(b)=f(1)=(1)^{3}-2(1)^{3}-1+3\\ &=1-2-1+3\\ &=4-3\\ &=1\\ &\text { Now, let us show that there exist } c \in(0,1) \text { such that }\\ &f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \end{aligned}

f(x)=x^{3}-2 x^{2}-x+3
On differentiating above with respect to x, we get
f^{\prime}(x)=3 x^{2}-4 x-1
Put x=c in above equation, we get
f^{\prime}(c)=3 c^{2}-4 c-1 \ldots(i)
By Mean Value Theorem,
\\f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$ \\$f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}$ \\$\Rightarrow f^{\prime}(c)=\frac{1-3}{1-0}$ \\$\Rightarrow f^{\prime}(c)=-\frac{2}{1}

\begin{aligned} &\Rightarrow f^{\prime}(c)=-2\\ &\Rightarrow 3 c^{2}-4 c-1=-2[\text { from }(j)]\\ &\begin{array}{l} \Rightarrow 3 c^{2}-4 c-1+2=0 \\ \Rightarrow 3 c^{2}-4 c+1=0 \end{array}\\ &\text { On factorising, we get }\\ &\Rightarrow 3 c^{2}-3 c-c+1=0\\ &\begin{array}{l} \Rightarrow 3 c(c-1)-1(c-1)=0 \\ \Rightarrow(3 c-1)(c-1)=0 \\ \Rightarrow(3 c-1)=0 \text { or }(c-1)=0 \end{array} \end{aligned}

\begin{aligned} &\Rightarrow c=\frac{1}{3} \text { or } c=1\\ &\text { So, value of }\\ &c=\frac{1}{3} \in(0,1) \end{aligned}

Thus, Mean Value Theorem is verified.

Posted by

infoexpert22

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