Question

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

Answer 1

f(x) = (x – 1)e^x + 1, x > 0

f'(x) = (x – 1)e^x + e^x(1 – 0) + 0

= xe^x – e^x + e^x

= xe^x

For all x > 0

x.e^x > 0.e⁰

xe^x > 0

f'(x) > 0

Hence, function f(x) = (x – 1)e^x + 1 is an increasing function for all x > 0.