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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)
(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)
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