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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 ___

asked in Calculus | 213 views

+1 vote

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

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

Graph of F(1) from the obtained equation

1<F(1)<= 39/7
Thus, F(1)⋳(1,39/7)

answered by anonymous   1 1 2
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