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 evaluate the second derivative at the points of interest:

1. We already know stationary points occur at x = 0 and x = -4.

2. Evaluate f''(x) at these points:

   • For x = 0:

     $f''(0) = 12(0)^2 + 30(0) = 0$

     Here, we check values around x = 0:

     • For x = 1, $f''(1) = 12(1)^2 + 30(1) = 42 > 0$ (indicating a local minimum)
     • For x = -1, $f''(-1) = 12(-1)^2 + 30(-1) = -18 < 0$ (indicating a local maximum)

   • For x = -4:

     $f''(-4) = 12(-4)^2 + 30(-4) = 0$

     Again, check values around x = -4:

     • For x = -3, $f''(-3) = 12(-3)^2 + 30(-3) = 12 > 0$ (indicating a local minimum)
     • For x = -5, $f''(-5) = 12(-5)^2 + 30(-5) = -18 < 0$ (indicating a local maximum)

3. Hence, we conclude:

   • At x = 0, there is a point of inflection.
   • At x = -4, this suggests a nature change, confirming a stationary point.

To determine where the function is increasing, we need f'(x) to be greater than or equal to zero:

1. Set up the inequality:

   $f'(x) = 4x^3 + 15x^2 \\ 0$

2. Factor the derivative:

   $f'(x) = x^2(4x + 15) \\ 0$

3. Analyze the critical points:

   • Critical point at x = 0 (stationary point)
   • The expression is zero when x = 0 or x = -rac{15}{4}.

4. Test intervals to find where the function is increasing:

   • For x < -rac{15}{4}, f'(x) < 0.
   • For -rac{15}{4} < x < 0, f'(x) > 0.
   • For x > 0, f'(x) > 0.

5. Thus, the function is increasing for:

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

The function g is defined as:

$g(x) = x^4 - 5x^3$

1. To find the transformation:
   • Notice that g(x) can be seen as a reflection of f(x) across the x-axis, as all terms of f(x) have been multiplied by -1:
   $g(x) = -f(x)$

To determine where g is increasing, we evaluate g'(x):

1. First, find g'(x):

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

2. Set up the inequality for increasing intervals:

   $g'(x) = 4x^2(x - \frac{15}{4}) > 0$

3. Analyze the critical points:

   • Critical point occurs at x = 0 and x = \frac{15}{4}.

4. Test intervals for increase:

   • For x < 0, g'(x) < 0.
   • For 0 < x < \frac{15}{4}, g'(x) > 0.
   • For x > \frac{15}{4}, g'(x) < 0.

5. Thus, g is increasing in the range:

   $0 < x < \frac{15}{4}$