# Q help (1 Viewer)

#### ExtremelyBoredUser

##### hiiii
View attachment 35309
Hey, can somene help me with this question and in the working out please explain why each step was done because my calculus skills are a little bad, thanks in advanced
i)
This is the same as showing for x > 0. Let h(x) = e^x - 1 -x

for x > 0
hence for x>0;

therefore is positive for x > 0 and this implies is solely increasing for x > 0

Therefore h(x) > 0 for x > 0 ( as h(x) is increasing for domain (x,infinity) and has a range of (0, infinity) )

ii)

to show a function is concave up for all x,

Hence f(x) is concave up for all x as f''(x) > 0 for all x

iii)

This is similar to the reasoning in 1), as it is proven that the second derivative is increasing for all x so now we just have to check f'(x) at x = 0

Hence for x > 0 as f''(x) is increasing for all values of x and f'(x) has range (0,infty) given x > 0

Now this is just the same process for f(x).

Therefore f(x) > 0 for x > 0 as f'(x) is increasing for x > 0 and f(x) has range (0,infty) given x > 0

f(x) > 0 so;