Show that f(x) = (x – 1) e^{x} + 1 is an increasing function for all x > 0.

Given:- Function f(x) = (x – 1) e^{x} + 1

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 – 1) e^{x} + 1

⇒

⇒ f’(x) = e^{x} + (x – 1) e^{x}

⇒ f’(x) = e^{x}(1+ x – 1)

⇒ f’(x) = xe^{x}

as given

x > 0

⇒ e^{x} > 0

⇒ xe^{x} > 0

⇒ f’(x) > 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval x > 0

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