As described in our section Definition of the Derivative - First Principles, the first derivative of the function \( f(x) \) which we write as \( f'(x) \) or as \( \frac{dy}{dx} \), is the slope of the tangent line to the function at the point \( x \). To put this in non-graphical terms, the first derivative tells us how whether a function is increasing or decreasing, and by how much it is increasing or decreasing.
This information is reflected in the graph of a function by the slope of the tangent line to a point on the graph, which is sometimes describe as the slope of the function. Positive slope tells us that, as \( x \) increases, \( f(x) \) also increases. Negative slope tells us that, as \( x \) increases, \( f(x) \) decreases. A zero slope does not tell us anything in particular; the function may be increasing, decreasing, or at a local maximum or a local minimum at that point. Writing this information in terms of derivatives, we see that:
If \( f'(p) > 0 \), then \( f(x) \) is an increasing function at \( x = p \);
If \( f'(p) < 0 \), then \( f(x) \) is a decreasing function at \( x = p \);
If \( f'(p) = 0\) , then \( x = p \) is called a critical point of \( f(x) \), and we do not know anything new about the behavior of \( f(x) \) at \( x = p \).
Consider \( f(x) = 3x^3 − 6x^2 + 2x − 1 \). We then have \( f'(x) = 9x^2 − 12x + 2 \).
At \( x = 0, f'(x) = 2 > 0\), so we know that \( f(x) \) is increasing at \( x = 0 \);
At \( x = 1, f'(x) = −1 < 0 \), so \( f(x) \) is decreasing at \( x = 1 \).
We have discussed how to determine whether a function is increasing or decreasing at a certain point without drawing a graph. This concept can be taken a step further to determine our intervals of increase and decrease. If we can determine all of the values where the derivative is positive, negative or zero, we can eliminate solving for \( x \) one number at a time.
Consider \( f(x) = \frac{2}{3}x^3 - 3x^2 - 8x + 7 \). Applying our differentiation rules, it follows that \( f'(x) = 2x^2 - 6x - 8 \).
\( \Longrightarrow \) Let's determine the x-intercepts.
\( \Longrightarrow 2x^2 - 6x - 8 = 2(x^2 - 3x - 4) = 2(x-4)(x+1) \)
\( \Longrightarrow x = -1, 4 \)
Our first step to determining our intervals of increase/decrease is to create a table like the one below:
\( ( -\infty, -1) \) | \( -1 \) | \( (-1, 4) \) | \( 4 \) | \( (4, \infty) \) | |
\( f'(x) \) | \( 0 \) | \( 0 \) | |||
\( f(x) \) |
We will now determine where \( f'(x) \) is positive and negative. Since we only have two x-intercepts, we know that the first derivative can only be either below or above the \( x-axis \) in the intervals of \( (-\infty, -1), (-1, 4), (4, \infty) \). It follows that any value within those intervals can be used to determine where the derivative lies within that interval.
\( \Longrightarrow -2 \in (-\infty, -1) \rightarrow f'(-2) = 12 > 0 \)
\( \Longrightarrow 0 \in (-1, 4) \rightarrow f'(0) = -8 < 0 \)
\( \Longrightarrow 5 \in (4, \infty) \rightarrow f'(5) = 12 > 0 \)
With this information, we are able to complete the table and determine where \( f(x) \) is increasing and decreasing:
\( ( -\infty, -1) \) | \( -1 \) | \( (-1, 4) \) | \( 4 \) | \( (4, \infty) \) | |
\( f'(x) \) | + | \( 0 \) | - | \( 0 \) | + |
\( f(x) \) | \( increasing \) | \( decreasing \) |
\( increasing \) |
EXAMPLE
Consider \( f(x) = 4x^4 + 2x^3 - x^2 \). Determine the intervals of increase and decrease of the function using the first derivative.
See the video below for the solution: