Question

Prove that f(x)=(x-1)eˣ+1 is increasing for all x>0.

Answer 1

function , f(x) = (x - 1)e^x + 1

differentiating f(x) with respect to x,

f'(x) = d[(x - 1)e^x + 1]/dx

= d[(x - 1)e^x]/dx + d(1)/dx

= e^x d(x - 1)/dx + (x - 1) d(e^x)/dx + 0

= e^x × 1 + (x - 1) × e^x

= e^x + (x - 1)e^x

= e^x(x - 1 + 1)

= xe^x

hence, f'(x) = xe^x

for all x > 0

⇒e^x > e^0

⇒ e^x > 1 > 0

so, f'(x) = xe^x > 0

we know, any function f is increasing in (a,b) only when f' > 0 in interval (a, b).

here, f'(x) > 0 in interval, x > 0

so, f(x) is increasing for all x > 0