Lagrange 's Mean Value Theorem 2

Question

If f ' (x) =$\frac{8}{x^{}2+3x+4}$ and f(0) =1 then the lower and upper bounds of f(1) estimated by Langrange 's Mean Value Theorem are ___

Answer 1

$8/(x^{2}+3x+4) =f(1) - f(0)/ (1-0)$

=> $8/(x^{2}+3x+4) =f(1) - 1$
=> $(x^{2}+3x+12)/(x^{2}+3x+4) =f(1)$

Graph of F(1) from the obtained equation

1<F(1)<= 39/7
Thus, F(1)⋳(1,39/7)

Comments

how can it be 1 for lowe bound..??

how did u find upper bound?if it dun want to make a graph,then how to find?by maxima minima i am getting 39/7

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How lower and upper bound for f(1) is calculated....???

Answer 2

By Cauchy's mean value theorem,
f'(c)=(f(1)-f(0))/(1-0) ,  where c belongs to (0,1)

f(1)=f(0)+f'(c)
       =1+8/(c^2+3c+4)
       =(c^2+3c+12)/(c^2+3c+4) ,

           where c belongs to(0,1)

To find upper and lower bounds of f(1), putting lim(c->0) and lim(c->1) respectively,

lim(c-->0) f(1)=3 (upper bound)
lim(c-->1) f(1)=2 (lower bound)

Thanks !!!

Comments

What made you think to use limit to evaluate lower bound and upper bound