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 analyze the nature of the stationary points, we evaluate the second derivative at the stationary points found earlier, which are x = 0 and x = -4.

1. At x = 0: $f''(0) = 12(0)^2 + 30(0) = 0$ Since the second derivative is zero, we need to consider the first derivative:

   • Choose points to the left and right of 0 (e.g., x = -1 and x = 1):
   • $f'(-1) = 4(-1)^3 + 15(-1)^2 = -4 + 15 = 11 > 0$ (increasing)
   • $f'(1) = 4(1)^3 + 15(1)^2 = 4 + 15 = 19 > 0$ (increasing) Hence, x = 0 is a point of inflection.

2. At x = -4: $f''(-4) = 12(-4)^2 + 30(-4) = 192 - 120 = 72$ (positive) Therefore, x = -4 is a local minimum.

In conclusion, we have determined the nature of the stationary points: x = 0 is a point of inflection, and x = -4 is a local minimum.

For the function f(x) to be increasing, its first derivative must be greater than or equal to zero: $f'(x) = 4x^3 + 15x^2 ext{ must be } extgreater 0$.

Factoring gives: $f'(x) = x^2(4x + 15)$.

The critical points occur at:

1. x = 0
2. 4x + 15 = 0 → x = -rac{15}{4}

Analyzing the intervals:

• For x < -rac{15}{4}, both factors are negative, so f'(x) < 0.
• For -rac{15}{4} < x < 0, the first factor is positive and the second factor is negative, so f'(x) < 0.
• For x > 0, both factors are positive, yielding f'(x) > 0.

Thus, the range of values where f(x) is increasing is: $x > 0$.

The function g(x) is obtained by reflecting f(x) across the x-axis. Therefore, the transformation that maps f onto g is a reflection in the x-axis.

To find when g(x) is increasing, we start with: $g(x) = -x^4 - 5x^3.$

Calculating the first derivative: $g'(x) = -4x^3 - 15x^2.$

Setting the derivative greater than zero for increasing behavior: $g'(x) < 0$

Factoring gives: $g'(x) = -x^2(4x + 15) < 0$

Which has critical points:

1. x = 0
2. 4x + 15 = 0 → x = -rac{15}{4}

Analyzing the intervals:

• For x < -rac{15}{4}, both factors are positive, hence g'(x) > 0 (increasing).
• For -rac{15}{4} < x < 0, g'(x) < 0 (decreasing).

Therefore, g(x) is increasing for: x < -rac{15}{4}.