By function behavior, we mean the rate of change.

We have two flavors of this.

Each is a different measurement of the change in a function’s value compared to changes in the domain.

Algebra describes the change from one domain number to another. It describes function change over an interval. Calculus takes an extreme viewpoint on this. Calculus asks about the interval \([a,a]\). Algebra doesn’t know what to do with this type of interval. The rate of change of a function at one domain number, \(a\), makes no sense, algebraically. Calculus makes sense of it as the slope of a tangent line at \((a, f(a))\).

This gives us two types of interpretations for increasing and decreasing.

\(\blacktriangleright \) Algebraic Increasing

The function, \(f\), is increasing on the interval \([a, b]\) if

\(\blacktriangleright \) Calculus Increasing

The function, \(f\), is increasing at a if

\(\blacktriangleright \) Algebraic Decreasing

The function, \(f\), is decreasing on the interval \([a, b]\) if

\(\blacktriangleright \) Calculus Decreasing

The function, \(f\), is decreasing at a if

The instantaneous rate of change can have a value at each domain number, which makes it into a function.

iRoC

The instantaneous rate of change (a.k.a the derivative) gives us a formula for the slopes of tangent lines, which are rates of change.

\(\blacktriangleright \) From vertex form, \(f(x) = a (x -h)^2 + k\), we have \(iRoC_f(x) = f'(x) = 2 a \, (x-h)\).

\(\blacktriangleright \) From standard form, \(f(x) = a \, x^2 + b \, x + c\), we have \(iRoC_f(x) = f'(x) = 2 a \, x + b\).

Let \(M(t) = -\frac {1}{2} (t - 3)^2 + 5\)

What is the maximum value of \(M(t)\)?

From the vertex form, we can see a negative leading coefficient, which tells us the parabola will open down, that \(M\) increases and then decreases, and that there is a maximum value.

From the vertex form, we can tell that the highest point on the graph is \((3, 5)\), which tells us that the maximum value of \(M\) is \(5\) and it occurs at \(3\).

Let \(M(t) = -\frac {1}{2} (t - 3)^2 + 5\)

What is the maximum value of \(M(t)\)?

From the vertex form, we have \(iRoC_M(t) = -(t - 3)\) or \(M'(t) = -(t - 3) = -t + 3 = 3 - t\).

The \(iRoC\) or derivative is itself a function. We could plot its graph. For a quadratic function, the derivative is a linear function whose graph is a line.

Graph of \(y = iRoC_M(t) = M'(t) = 3 - t\)

The derivative is positive on \((-\infty , 3)\), which tells us that the tangent lines to the parabola have positive slopes for the left half of the parabola. This tells us that \(M\) is increasing decreasing on \((-\infty , 3)\).

The derivative is negative on \((3, \infty )\), which tells us that the tangent lines to the parabola have negative slopes for the right half of the parabola. This tells us that \(M\) is increasing decreasing on \((3, \infty )\).

When the derivative equals \(0\), we have a critical number.

\(3 - t = 0\) when \(t=3\)

Since, \(M\) increases and then decreases, we know this critical number marks the location of a maximum value of \(M\). To get this maximum value, we evaluate \(M\) at the critical number: \(M(3) = 5\).

\(M\) has a maximum value of \(5\), which occurs at \(3\).

For our example, \(M'(t) = -(t - 3) = -t + 3 = 3 - t\) is a function. The values of this function give the slopes of the tangent lines to the graph of \(M\).

For Instance:

The point \((1, 3)\) is on the graph of \(M\). The slope of the tangent line at \((1, 3)\) is \(M'(1) = 3 - 1 = 2\). Therefore, the equation of the tangent line is \(y - 3 = 2(t-1)\).