A function f is defined for all real values of x as f(x) = x^4 + 5x^3 The function has exactly two stationary points when x = 0 and x = -4 - AQA - A-Level Maths Mechanics - Question 9 - 2021 - Paper 3

To analyze the nature of the stationary points at x = 0 and x = -4, we follow these steps:

1. Calculate f''(0) and f''(-4):

   • For x = 0: $f''(0) = 12(0)^2 + 30(0) = 0$
   • For x = -4: $f''(-4) = 12(-4)^2 + 30(-4) = 12(16) - 120 = 192 - 120 = 72$

2. The second derivative at x = -4 is positive (f''(-4) > 0), indicating a local minimum at this stationary point.

3. The second derivative at x = 0 is zero, requiring further investigation. To do this, we can check the sign of f''(x) around x = 0:

   • For a point slightly less than 0 (e.g. x = -1): $f''(-1) = 12(-1)^2 + 30(-1) = 12 - 30 = -18 \quad (\text{negative})$
   • For a point slightly greater than 0 (e.g. x = 1): $f''(1) = 12(1)^2 + 30(1) = 12 + 30 = 42 \quad (\text{positive})$

Since f''(x) changes from negative to positive, this indicates that there is a point of inflection at x = 0.

To find the range of values where f(x) is increasing, we need to determine where f'(x) is greater than or equal to zero:

1. Set the first derivative f'(x) to zero: $4x^3 + 15x^2 = 0$

2. Factor the equation: $x^2(4x + 15) = 0$ This gives the stationary points at: x = 0 and x = -\frac{15}{4}.

3. Analyze the intervals formed by these points to determine where f'(x) is positive. We find that f(x) is increasing:
   $x \in \left[-\frac{15}{4}, \infty \right)$

The transformation that maps function f onto function g involves reflecting the graph of f in the y-axis. Thus, we write:

• The transformation is a reflection in the y-axis.

To find where g(x) is increasing, we again look for where g'(x) is greater than or equal to zero. First, we find the derivative:

1. Differentiate g(x): $g'(x) = -f'(x) = -(4x^3 + 15x^2)$

2. Set g'(x) to zero: $- (4x^3 + 15x^2) = 0 \Rightarrow 4x^3 + 15x^2 = 0$

3. Factor: $x^2(4x + 15) = 0 \quad \Rightarrow x=0 \text{ or } x=-\frac{15}{4}$

4. Test intervals. The function g(x) will be increasing when x is greater than these stationary points: $x \in \left(-\infty, -\frac{15}{4} \right]$. Therefore, the range of values x for which g is increasing is:
   $x \in \left(-\infty, -\frac{15}{4} \right)$.