Show that f(x) = x^{3} – 15x^{2} + 75x – 50 is an increasing function for all x ϵ R.

Given:- Function f(x) = x^{3} – 15x^{2} + 75x – 50

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^{3} – 15x^{2} + 75x – 50

⇒

⇒ f’(x) = 3x^{2} – 30x + 75

⇒ f’(x) = 3(x^{2} – 10x + 25)

⇒ f’(x) = 3(x – 5)^{2}

Now, as given

x ϵ R

⇒ (x – 5)^{2} > 0

⇒ 3(x – 5)^{2} > 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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