Using the given formula for the \(n^{\text{th}}\) derivative, \(f^{(n)}(x) = (2^n x + n 2^{n-1}) e^{2x}\):

For \(n = 1\): \(f'(x) = (2^1 x + 1 \cdot 2^{1-1}) e^{2x} = (2x + 1) e^{2x}\) A1

Set \(f'(x) = 0\) to find critical points: \((2x + 1) e^{2x} = 0\) M1

Since \(e^{2x} \neq 0\), \(2x + 1 = 0 \implies x = -\frac{1}{2}\) A1

For \(n = 2\): \(f”(x) = (2^2 x + 2 \cdot 2^{2-1}) e^{2x} = (4x + 4) e^{2x}\) A1

Evaluate at \(x = -\frac{1}{2}\): \(f”\left(-\frac{1}{2}\right) = \left(4 \cdot \left(-\frac{1}{2}\right) + 4\right) e^{2 \cdot (-\frac{1}{2})} = (4 \cdot (-\frac{1}{2}) + 4) e^{-1} = (-2 + 4) \cdot \frac{1}{e} = \frac{2}{e} > 0\) A1

Since \(f”\left(-\frac{1}{2}\right) > 0\), the point at \(x = -\frac{1}{2}\) is a local minimum. R1

Find coordinates of P: \(f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right) e^{2 \cdot (-\frac{1}{2})} = -\frac{1}{2} e^{-1} = -\frac{1}{2e}\)

Thus, P is \(\left(-\frac{1}{2}, -\frac{1}{2e}\right)\) A1

[7 marks]

A point of inflection occurs where \(f”(x) = 0\) and the concavity changes.

From part (a), \(f”(x) = (4x + 4) e^{2x}\). Set \(f”(x) = 0\):

\((4x + 4) e^{2x} = 0 \implies 4x + 4 = 0 \implies x = -1\) (since \(e^{2x} \neq 0\)) M1A1

Check concavity change using \(f”(x)\):

At \(x = -\frac{1}{2}\): \(f”\left(-\frac{1}{2}\right) = \frac{2}{e} > 0\) (concave up)

At \(x = -2\): \(f”(-2) = (4 \cdot (-2) + 4) e^{2 \cdot (-2)} = (-8 + 4) e^{-4} = -\frac{4}{e^4} < 0\) (concave down) M1A1

The sign change of \(f”(x)\) around \(x = -1\) indicates a point of inflection at \(x = -1\). R1

[5 marks]

Concavity is determined by the sign of \(f”(x) = (4x + 4) e^{2x}\). Since \(e^{2x} > 0\), the sign depends on \(4x + 4\).

(i) Concave up when \(f”(x) > 0\): \(4x + 4 > 0 \implies x > -1\) A1

(ii) Concave down when \(f”(x) < 0\): \(4x + 4 < 0 \implies x < -1\) A1

[2 marks]

Key features for the sketch:

– Intercepts: At \(x = 0\), \(f(0) = 0 \cdot e^0 = 0\), so \((0, 0)\) A1

– Asymptote: As \(x \to -\infty\), \(e^{2x} \to 0\), so \(f(x) \to 0\), giving a horizontal asymptote at \(y = 0\). As \(x \to \infty\), \(f(x) \to \infty\) A1

– Points: P at \(\left(-\frac{1}{2}, -\frac{1}{2e}\right)\) (minimum), Q at \((-1, f(-1)) = (-1, -e^{-2}) = (-1, -\frac{1}{e^2})\) (inflection) A1

– Shape: Concave down for \(x < -1\), concave up for \(x > -1\), minimum at P, inflection at Q A1

[4 marks]

Prove by mathematical induction that \(f^{(n)}(x) = (2^n x + n 2^{n-1}) e^{2x}\) for all \(n \in \mathbb{Z}^+\).

Base case (\(n = 1\)):

\(f(x) = x e^{2x}\), \(f'(x) = e^{2x} + x \cdot 2 e^{2x} = e^{2x} (1 + 2x) = (2x + 1) e^{2x}\)

For \(n = 1\): \(f^{(1)}(x) = (2^1 x + 1 \cdot 2^{1-1}) e^{2x} = (2x + 1) e^{2x}\), which holds. M1A1

Inductive step: Assume true for \(n = k\): \(f^{(k)}(x) = (2^k x + k 2^{k-1}) e^{2x}\). M1A1

Show for \(n = k+1\): Compute \(f^{(k+1)}(x) = \frac{d}{dx} \left( f^{(k)}(x) \right)\):

\(f^{(k)}(x) = (2^k x + k 2^{k-1}) e^{2x}\)

\(\frac{d}{dx} \left( (2^k x + k 2^{k-1}) e^{2x} \right) = 2^k e^{2x} + (2^k x + k 2^{k-1}) \cdot 2 e^{2x}\)

\(= e^{2x} (2^k + 2 (2^k x + k 2^{k-1})) = e^{2x} (2^k + 2^{k+1} x + k 2^k)\)

\(= e^{2x} (2^{k+1} x + 2^k + k 2^k) = e^{2x} (2^{k+1} x + (k+1) 2^k)\) A1A1

This matches the form for \(n = k+1\). Since true for \(n = 1\) and assuming true for \(n = k\) implies true for \(n = k+1\), the result holds for all \(n \in \mathbb{Z}^+\) by induction. R1

[9 marks]

Total [27 marks]