• Find derivative: \( f_1′(x) = e^{-x} – xe^{-x} = (1-x)e^{-x} \)

• Set \( f_1′(x) = 0 \Rightarrow x = 1 \)

• Maximum at \( (1, e^{-1}) \approx (1, 0.368) \)

• Graph passes through \( (0,0) \) and approaches \( 0 \) as \( x \to \infty \)

\( \int_0^b xe^{-x}dx = \left[-xe^{-x}\right]_0^b + \int_0^b e^{-x}dx \)

\( = -be^{-b} + \left[-e^{-x}\right]_0^b \)

\( = -be^{-b} + (-e^{-b} + 1) \)

\( = 1 – e^{-b} – be^{-b} \)

\( = \frac{e^b – 1 – b}{e^b} \)

\( \lim_{b \to \infty} \frac{e^b – b – 1}{e^b} \)

Apply l’Hôpital’s rule (\( \frac{\infty}{\infty} \) form):

\( = \lim_{b \to \infty} \frac{e^b – 1}{e^b} \)

Apply again:

\( = \lim_{b \to \infty} \frac{e^b}{e^b} = 1 \)

\( A_1 = \lim_{b \to \infty} \) of part (b) result \( = 1 \)

(i) \( A_4 \) [2 marks]

Using GDC: \( \int_0^\infty x^4e^{-x}dx \approx 24 \)

(ii) \( A_5 \) [2 marks]

Using GDC: \( \int_0^\infty x^5e^{-x}dx \approx 120 \)

Observing pattern:

\( A_1 = 1 = 1! \), \( A_2 = 2 = 2! \), \( A_3 = 6 = 3! \), \( A_4 = 24 = 4! \), \( A_5 = 120 = 5! \)

Conjecture: \( A_n = n! \)

Base Case (\( n=1 \)): Verified \( A_1 = 1! \)

Inductive Step:

Assume \( A_k = k! \) holds

For \( A_{k+1} = \int x^{k+1}e^{-x}dx \):

\( = \left[-x^{k+1}e^{-x}\right]_0^\infty + (k+1)\int x^ke^{-x}dx \)

\( = 0 + (k+1)A_k \)

\( = (k+1)k! = (k+1)! \)

Thus true for \( n=k+1 \) when true for \( n=k \)