The functions f and g are defined by f: x ↦ 2x + ln 2, x ∈ ℝ, g: x ↦ e^{2x}, x ∈ ℝ - Edexcel - A-Level Maths Pure - Question 1 - 2006 - Paper 5

To sketch the curve y = gf(x) = 4e^{4x}, we note that:

1. When x = 0, the value of y is: $y = gf(0) = 4e^{4(0)} = 4e^0 = 4.$

Thus, the coordinates where the curve cuts the y-axis are (0, 4).

2. The curve is an exponential function that increases rapidly as x increases.

The function gf(x) = 4e^{4x} is an exponential function.

1. Since the exponential function $e^{4x}$ is always positive, the range is: [ (0, +\infty) ] Therefore, the range of gf is ( (0, +\infty) ).

To find ( \frac{d}{dx}[gf(x)] ):

1. First, compute the derivative: $gf(x) = 4e^{4x}$ Using the chain rule: $\frac{d}{dx}[gf(x)] = 4 \cdot 4e^{4x} = 16e^{4x}$

2. Set this equal to 3: $16e^{4x} = 3$

3. Solving for x gives: $e^{4x} = \frac{3}{16}$

4. Taking the natural logarithm on both sides: $4x = ln(\frac{3}{16})$ $x = \frac{1}{4}ln(\frac{3}{16})$

5. Evaluating this using a calculator: $x \approx -0.418.$

Thus, the value of x is approximately -0.418 to 3 significant figures.