Given the function: $$f(x) = x^4 - 4x - 8$$ (a) Show that there is a root of $f(x)=0$ in the interval $[-2, -1]$ - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 6

To find the turning point, we first compute the first derivative: $f'(x) = 4x^3 - 4$

Setting the derivative to zero to find critical points: $4x^3 - 4 = 0 \implies x^3 = 1 \implies x = 1$

Next, we find the $y$-coordinate by evaluating $f(1)$: $f(1) = 1^4 - 4(1) - 8 = 1 - 4 - 8 = -11$

Thus, the coordinates of the turning point are $(1, -11)$.

Expanding the given form: $f(x) = (x-2)(x^2 + ax^2 + bx + c) = (x - 2)(x^2 + ax^2 + bx + c) = (x - 2)(1 + a)x^2 + (b - 2c)x - 2c$

Matching coefficients with $f(x) = x^4 - 4x - 8$, we have:

• For $x^3$: $1 + a = 0 \implies a = -1$
• For $x^2$: $b - 2c = -4$
• For the constant term: $-2c = -8 \implies c = 4$ Thus, substituting $c$ back, we find $b - 8 = -4 \implies b = 4$. Therefore, $a = -1, b = 4, c = 4$.

The graph of $y = f(x)$ will have the following features:

• It intersects the $y$-axis at $-8$ (when $x=0$).
• The turning point at $(1, -11)$ is where the graph has a minimum, suggesting the graph decreases until $x=1$ and then increases afterwards.
• Roots are in the interval $[-2, -1]$ as identified earlier. The sketch would show this basic shape with appropriate curvature near the turning point and crossing the axis.

To sketch $y = |f(x)|$, reflect any part of the graph of $y = f(x)$ that is below the $x$-axis to above the $x$-axis. This implies:

• The portion of the graph between the roots will be reflected.
• The turning point $(1, -11)$ will be reflected to $(1, 11)$.

The overall shape of $y = |f(x)|$ will resemble the original graph but with all negative values flipped above the $x$-axis.