# Q help (1 Viewer)

#### Anounymouse352

##### New Member

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

#### 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 $\bg_white e^x - 1 - x > 0$ for x > 0. Let h(x) = e^x - 1 -x

$\bg_white h'(x) = e^x - 1$
for x > 0 $\bg_white e^x > 1$
hence for x>0;
$\bg_white e^x - 1 > 0$

therefore $\bg_white h'(x)$ is positive for x > 0 and this implies $\bg_white h(x)$ is solely increasing for x > 0

$\bg_white h(0) = e^0 - 1 - 0 = 1 - 1 - 0 = 0$

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, $\bg_white f''(x) > 0$

$\bg_white f(x) = e^x - 1 - x + x^2$
$\bg_white f'(x) = e^x - 1 + 2x$
$\bg_white f''(x) = e^x + 2$

$\bg_white e^x > 0 \Rightarrow e^x + 2 > 2 > 0$

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

$\bg_white f'(0) = e^0 - 1 + 2(0) = 1 - 1 = 0$

Hence $\bg_white f'(x) > 0$ 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).

$\bg_white f(0) = e^0 - 1 - 0 + 0 = 0$

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;

$\bg_white e^x - 1 - x + x^2 > 0$
$\bg_white e^x > 1 + x - x^2$