Sketching Gradient Functions (Edexcel A-Level Mathematics): Revision Notes

• Where $\ f(x)$ is increasing: $\ f'(x) > 0$ (positive slope). The graph of $\ f'(x)$ lies above the $x$-axis.
• Where $\ f(x)$ is decreasing: $\ f'(x) < 0$ (negative slope). The graph of $\ f'(x)$ lies below the $x$-axis.
• Where $f(x)$ has a stationary point: $\ f'(x) = 0 .$ The graph of $\ f'(x)$ crosses the $x$-axis.
• A local maximum of $\ f(x)$ corresponds to $\ f'(x)$ crossing the $x$-axis from positive to negative.
• A local minimum of $\ f(x)$ corresponds to $\ f'(x)$ crossing the $x$-axis from negative to positive.

• Where $\ f(x)$ is concave up: $\ f'(x)$ is increasing.
• Where $\ f(x)$ is concave down: $\ f'(x)$ is decreasing.

Step 1: Analyse the Function $\ f(x)$

• Identify key features of the original function $\ f(x) :$
• Stationary points: Where $\ f'(x) = 0$ (local maxima, minima, or points of inflection).
• Intervals where $\ f(x)$ is increasing or decreasing: Identify where$\ f'(x)$ is positive or negative.
• Concavity: Determine where $\ f(x)$ is concave up or concave down to understand the shape of $\ f'(x)$.

Step 2: Determine the Signs of $\ f'(x)$

• Positive slope: Sketch $\ f'(x)$ above the x-axis where $\ f(x)$ is increasing.
• Negative slope: Sketch $\ f'(x)$ below the x-axis where $\ f(x)$ is decreasing.
• Zero slope: Mark points where $\ f(x)$ has a stationary point (crosses the $x$-axis).

Step 3: Draw the General Shape of $\ f'(x)$

• Crossing the $x$-axis: At each stationary point of $\ f(x)$, the gradient function $\ f'(x)$ will cross the $x$-axis.
• Curvature: Consider the concavity of $\ f(x)$:
• If $\ f(x)$ is concave up (curving upwards), $\ f'(x)$ is increasing.
• If $\ f(x)$ is concave down (curving downwards), $\ f'(x)$ is decreasing.
• Asymptotic behaviour: If $\ f(x)$ has asymptotes or tends to infinity, consider how $\ f'(x)$behaves as $\ x$ approaches those values.

Step 4: Sketch the Gradient Function $\ f'(x)$

• Combine the information about where $\ f'(x)$ is positive, negative, and zero.
• Draw the curve of $\ f'(x)$ smoothly, reflecting the nature of the slopes at various points of $\ f(x) .$

Example 1: Sketch the gradient function of $\ f(x) = x^2 - 4x + 3$

• Step 1: Analyse$\ f(x) :$
• $\ f(x) = x^2 - 4x + 3$ is a quadratic function, so its graph is a parabola.
• Find the derivative: $f'(x) = 2x - 4$
• The stationary point occurs where $\ f'(x) = 0$: $2x - 4 = 0 \quad \Rightarrow \quad x = 2$
• At $\ x = 2 \ , \ f(x) = 2^2 - 4(2) + 3 = 4 - 8 + 3 = -1 .$

• Step 2: Sketch $\ f'(x)$:
• $\ f'(x) = 2x - 4$ is a linear function with a slope of $2$, crossing the $x$-axis at $\ x = 2 .$
• For $\ x < 2 \ , \ f'(x) < 0$ (decreasing function).
• For $\ x > 2 \ , \ f'(x) > 0$ (increasing function).
• The graph of $\ f'(x)$ is a straight line with a positive slope, crossing the $x$-axis at $\ x = 2$.

Example 2: Sketch the gradient function of $\ f(x) = x^3 - 3x^2 + 2x$

• Step 1: Analyse $\ f(x)$:
• $\ f(x) = x^3 - 3x^2 + 2x$ is a cubic function.
• Find the derivative: $f'(x) = 3x^2 - 6x + 2$
• Find the stationary points by solving $\ f'(x) = 0$: $3x^2 - 6x + 2 = 0$ $x = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6} = 1 \pm \frac{\sqrt{3}}{3}$
• These roots give the points where the gradient function $\ f'(x)$ crosses the $x$-axis.

• Step 2: Sketch $f'(x)$:
• The quadratic function $\ f'(x) = 3x^2 - 6x + 2$ is a parabola opening upwards.
• The points where $\ f'(x)$ crosses the x-axis are the stationary points of $\ f(x) .$
• The graph of $\ f'(x)$ will be a parabola that crosses the $x$-axis at these roots and opens upwards.

• Sketching gradient functions involves translating the behaviour of the original function $\ f(x)$into a graph of its derivative $\ f'(x) .$
• By analysing where $\ f(x)$ is increasing, decreasing, and where it has stationary points, you can determine the general shape of $\ f'(x) .$
• Understanding how the original function behaves allows you to accurately sketch the derivative and gain insights into the function's overall behaviour.