Increasing and decreasing functions
Watch this video to learn about increasing and decreasing functions.
Below is the graph of a quadratic function, showing where the function is increasing and decreasing.
If we draw in the tangents to the curve, you will notice that if the gradient of the tangent is positive, then the function is increasing and if the gradient is negative then the function is decreasing.
To calculate the gradient of the tangents, we differentiate in order to substitute the relevant \(x\) value in. Therefore we can say that when:
- \(\frac{{dy}}{{dx}}\textgreater0\) (A positive gradient means the line slopes up to the right.)\(\to\)Function is increasing
- \(\frac{{dy}}{{dx}} = 0\)\(\to\)Function is stationary
- \(\frac{{dy}}{{dx}}\textless0\) (A negative gradient means the line slopes down to the right.)\(\to\)Function is decreasing
The turning point is the point on the curve when it is stationary. This is why you will see turning points also being referred to as stationary points.
Example 1
Is the curve \(y = {x^2} - 6x\) increasing or decreasing at the point \(x = 5?\)
Solution
\(\frac{{dy}}{{dx}} = 2x - 6\)
When \(x = 5\),
\(\frac{{dy}}{{dx}} = 2(5) - 6\)
\(= 10 - 6 = 4\)
Since \(\frac{{dy}}{{dx}}\textgreater0\) then the curve is increasing at the point \(x = 5\).
If we know where the turning point of the curve is, then we can use this to determine the nature (whether the graph is increasing/decreasing) to the left and right of the stationary point.
Example 2
Find the intervals in which the function \(y = {x^3} - 3{x^2} + 8\) is increasing and decreasing, where the stationary points are at \(x = 0\) and \(x = 2\).
Solution
\(\frac{{dy}}{{dx}} = 3{x^2} - 6x\)
In order to determine the shape of the curve, we use a table to display the nature to the left and right of the stationary points.
To find the nature to the left of the stationary point, you choose an \(x\) value slightly less than the stationary value. Similarly to find the nature to the right of the stationary point, you choose an \(x\) value slightly bigger.
Take each value of \(x\) in turn and substitute in to \(\frac{{dy}}{{dx}}\) to find out whether it is greater than, less than or equal to zero, in order to determine the shape of the curve.
\(\frac{{dy}}{{dx}} = 3{( - 1)^2} - 6( - 1) = 3 + 6 = 9\) (positive)
\(\frac{{dy}}{{dx}} = 3{(0)^2} - 6(0) = 0 - 0 = 0\) (stationary)
\(\frac{{dy}}{{dx}} = 3{(1)^2} - 6(1) = 3 - 6 = - 3\) (negative)
\(\frac{{dy}}{{dx}} = 3{(2)^2} - 6(2) = 12 - 12 = 0\) (stationary)
\(\frac{{dy}}{{dx}} = 3{(3)^2} - 6(3) = 27 - 18 = 9\) (positive)
Therefore in summary:
The function is increasing when \(x\textless0\) and \(x\textgreater2\).
The function is decreasing when \(0\textless x\textless2\).
