Show that f(x) = x^{9} + 4x^{7} + 11 is an increasing function for all x ϵ R.

Given:- Function f(x) = x^{9} + 4x^{7} + 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) = x^{9} + 4x^{7} + 11

⇒

⇒ f’(x) = 9x^{8} + 28x^{6}

⇒ f’(x) = x^{6}(9x^{2} + 28)

as given

x ϵ R

⇒ x^{6} > 0 and 9x^{2} + 28 > 0

⇒ x^{6}(9x^{2} + 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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