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Let the function, f: \left [ -7,0\right ]\rightarrow R be continuous on \left [ -7,0\right ] and differentiable on \left [ -7,0\right ]. If f(-7)=-3 and f{}'(x)\leq 2, for all x\epsilon (-7,0), then for all such functions f, f(-1)+f(0) lies in the interval : 
Option: 1 \left [ -6 ,20\right ]      
Option: 2 \left [ -\infty ,20\right ]      
Option: 3 \left [ -\infty ,11\right ]      
Option: 4 \left [ -3,11\right ]      
 

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}

-

Directly use

\\\frac{f(-1)-f(-7)}{-1+7}=f'(c)\;\;\;\;\;\&\;\;\;\;\;\;\;\frac{f(0)-f(-7)}{0+7}=f'(c)\\\\\frac{f(-1)-f(-7)}{-1+7}=f'(2)\;\;\;\;\;\&\;\;\;\;\;\;\;\frac{f(0)-f(-7)}{0+7}=f'(2)

we get

f(-1)\leq 9\;\;\;\;\;\;\&\;\;\;\;\;\;\;\;f(0)\leq 11

f(-1)+f(0)\leq 20

Correct Option (2)

Posted by

Kuldeep Maurya

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