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  • Let  be any function continuous on  and twice differentiable on 

Let f be any function continuous on [a,b] and twice differentiable on (a,b). If for all x\epsilon \left ( a,b \right ),f^{'}(x)>0 and f^{''}(x)<0, then for any c\epsilon (a,b),\frac{f(c)-f(a)}{f(b)-f(c)} is greater than : 
Option: 1 \frac{b-c}{c-a}
Option: 2 1
Option: 3 \frac{c-a}{b-c}
Option: 4 \frac{b+a}{b-a}
 

Answers (1)

best_answer

 

 

Lagrange’s Mean Value Theorem -

Lagrange’s Mean Value Theorem

Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions f defined on a closed interval [a, b] with f(a) = f(b) . The Mean Value Theorem generalized Rolle’s theorem by considering functions that do not necessarily have equal value at the endpoints. 

Statement

Let f (x) be a function defined on [a, b] such that

  1. it is continuous on [a, b],

  2. it is differentiable on (a, b).

Then there exists a real number c  (a, b) such that

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

Cauchy’s mean value Theorem

Cauchy's mean value theorem, also known as the extended mean value theorem. It states that if both function f(x) and g(x) are continuous on the closed interval [a, b] and differentiable on the open interval (a,  b) and g’(x) is not zero on that open interval, then there exists c in (a, b) such that

\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}

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Use LMVT for x \in[a,c]

\frac{\mathrm{f}(\mathrm{c})-\mathrm{f}(\mathrm{a})}{\mathrm{c}-\mathrm{a}}=\mathrm{f}^{\prime}(\alpha), \alpha \in(\mathrm{a}, \mathrm{c})

also use LMVT for x \in[c,b]

\frac{f(b)-f(c)}{b-c}=f^{\prime}(\beta), \beta \in(c, b)

\because f ''(x) < 0 \Rightarrow f '(x) is decreasing

\\ {f^{\prime}(\alpha)>f^{\prime}(\beta)} \\ {\frac{f(c)-f(a)}{c-a}>\frac{f(b)-f(c)}{b-c}} \\ {\frac{f(c)-f(a)}{f(b)-f(c)}>\frac{c-a}{b-c}(\because f(x)\text { is increasing) } }

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avinash.dongre

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