Using Differentiation to Determine Graph Shapes

The Role of Differentiation in Graphing

Differentiation provides a mathematical way to describe the 'shape' of a function without needing to plot every individual point. The first derivative, $f'(x)$, represents the gradient function, which tells us how steep the graph is at any specific value of $x$. By analysing the sign and value of $f'(x)$, we can determine the direction and curvature of the graph.

Increasing and Decreasing Functions

A function is said to be increasing on an interval if, as $x$ increases, the value of $y$ also increases. Visually, the graph is 'going uphill' from left to right. This occurs wherever the gradient is positive, so we look for regions where $f'(x) > 0$. Conversely, a function is decreasing on an interval if $y$ falls as $x$ increases, meaning the graph is 'going downhill'. This occurs where $f'(x) < 0$.

To find these intervals, you should:

1. Differentiate the function to find $f'(x)$.
2. Set up and solve the inequalities $f'(x) > 0$ and $f'(x) < 0$.
3. Use these intervals to sketch the rising and falling sections of the graph.

Finding Stationary Points

Stationary points are specific points on a curve where the gradient is zero. At these points, the tangent to the curve is horizontal. Stationary points often represent turning points where the graph changes direction. The ESAT specification focuses on identifying local maxima and local minima.

To find stationary points:

1. Find the derivative $f'(x)$.
2. Solve the equation $f'(x) = 0$ to find the $x$ coordinates of the stationary points.
3. Substitute these $x$ values back into the original function $y = f(x)$ to find the corresponding $y$ coordinates.

Determining the Nature of Stationary Points

Once you have found a stationary point, you must classify its 'nature' as either a local maximum or a local minimum. The most efficient method for the ESAT is the second derivative test:

• Local Maximum: If $f''(x) < 0$ at the stationary point, the gradient is decreasing (changing from positive to negative), creating a peak.
• Local Minimum: If $f''(x) > 0$ at the stationary point, the gradient is increasing (changing from negative to positive), creating a trough.

Worked Example

Determine the stationary points and the intervals where the function $f(x) = x^3 - 3x + 5$ is increasing or decreasing.

1. Differentiate: $f'(x) = 3x^2 - 3$.
2. Find stationary points: Set $3x^2 - 3 = 0$, which gives $x^2 = 1$, so $x = 1$ and $x = -1$.
3. Find coordinates: $f(1) = (1)^3 - 3(1) + 5 = 3$ and $f(-1) = (-1)^3 - 3(-1) + 5 = 7$. The points are $(1, 3)$ and $(-1, 7)$.
4. Check nature: $f''(x) = 6x$.
   • At $x = 1$, $f''(1) = 6 > 0$, so $(1, 3)$ is a local minimum.
   • At $x = -1$, $f''(-1) = -6 < 0$, so $(-1, 7)$ is a local maximum.
5. Identify intervals: $f'(x) > 0$ when $3x^2 - 3 > 0$, which occurs when $x < -1$ or $x > 1$. In these regions, the graph is increasing. The graph is decreasing for $-1 < x < 1$.

Connection to Earlier Topics

This calculus approach generalizes the rules for simpler functions. For a linear function $y = mx + c$, the derivative is $f'(x) = m$, confirming a constant gradient. For a quadratic $y = a(x + b)^2 + c$, the derivative is $f'(x) = 2a(x + b)$. Setting this to zero gives $x = -b$, which is the $x$ coordinate of the vertex (the stationary point) derived from transformation rules.