Solution — 2016 Question 3 Graphs| Differentiation & Applications| Integration & Applications

(a)

(b)

From GC, when $x = 0.5,$

\begin{matrix} Gradient \\ = \frac{d y}{d x} \\ = - 1.6065 \\ = - 1.61 ∎ (to 3 s.f.) \end{matrix}

(c)

When $x = 0.5,$

\begin{matrix} y & = e^{- x} - x^{2} \\ = e^{- (0.5)} - {0.5}^{2} \\ = 0.356 53 \end{matrix}

\begin{matrix} y - (- y_{1}) & = m (x - (- x_{1})) \\ y - (- (0.356 53)) & = - 1.6065 (x - (- (0.5))) \\ y + 0.356 53 & = - 1.6065 (x + 0.5) \\ = - 1.6065 x - 0.803 27 \\ y & = - 1.6065 x - 1.1598 ∎ \end{matrix}

(d)

\begin{matrix} \int_{0}^{k} (e^{- x} - x^{2}) d x \\ = {[- e^{- x} - \frac{x^{3}}{3}]}_{0}^{k} \\ = - e^{- k} - \frac{k^{3}}{3} - (- e^{- (0)} - \frac{0^{3}}{3}) \\ = - e^{- k} - \frac{k^{3}}{3} + 1 ∎ \end{matrix}