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 Pure - Question 9 - 2021 - Paper 3

To determine the nature of the stationary points, we need to evaluate f''(x) at the stationary points x = 0 and x = -4:

1. Evaluating at x = 0:

   • ( f''(0) = 12(0)^2 + 30(0) = 0 )
   • Since f''(0) = 0, we require further analysis. We can check values around x = 0 to see the sign of f''(x):
   • For x = 1: $f''(1) = 12(1)^2 + 30(1) = 12 + 30 = 42 > 0$ (concave up)
   • For x = -1:
     $f''(-1) = 12(-1)^2 + 30(-1) = 12 - 30 = -18 < 0$ (concave down)
   • Hence, there is a point of inflection at x = 0.

2. Evaluating at x = -4:

   • ( f''(-4) = 12(-4)^2 + 30(-4) = 192 - 120 = 72 > 0 )
   • Since f''(-4) > 0, the stationary point at x = -4 is a local minimum.

The function f(x) is increasing where its first derivative f'(x) is greater than or equal to zero:

$4x^3 + 15x^2 \geq 0$

Factoring gives:

$x^2(4x + 15) \geq 0$

The solutions occur when:

• x^2 = 0, which gives x = 0
• 4x + 15 = 0, which gives x = -\frac{15}{4}

The function is increasing for x in the range:

$x \geq -\frac{15}{4}$

The transformation that maps function f onto g is a reflection in the y-axis. This is because g(x) = f(-x), which indicates that every point of f at x is mirrored to g at -x.

For the function g(x), we determine where its first derivative g'(x) is non-negative:

$g'(x) = 4x^3 - 15x^2$

Setting g'(x) \geq 0 gives:

$x^2(4x - 15) \geq 0$

This factors to find critical points:

• x^2 = 0, gives x = 0
• 4x - 15 = 0, gives x = \frac{15}{4}

The function g is increasing for the range:

$x \leq 0 \text{ or } x \geq \frac{15}{4}$