Show that f(x) = x9 + 4x7 + 11 is an increasing function for all x ϵ R.

Given:- Function f(x) = x9 + 4x7 + 11


Theorem:- Let f be a differentiable real function defined on an open interval (a,b).


(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)


(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)


Algorithm:-


(i) Obtain the function and put it equal to f(x)


(ii) Find f’(x)


(iii) Put f’(x) > 0 and solve this inequation.


For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain, it is decreasing.


Here we have,


f(x) = x9 + 4x7 + 11



f’(x) = 9x8 + 28x6


f’(x) = x6(9x2 + 28)


as given


x ϵ R


x6 > 0 and 9x2 + 28 > 0


x6(9x2 + 28) > 0


f’(x) > 0


Hence, condition for f(x) to be increasing


Thus f(x) is increasing on interval x R


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