Rate of Change Over an Interval

Consider the quadratic function \(Q(x) = \frac {(x+7)(x-5)}{5}\)

The rate-of-change over the interval \([-3, 4]\) is

This is the slope of the secant line through the points \((-3, Q(-3))\) and \((4, Q(4))\).

This rate-of-change is the constant rate of change that would be needed in order for a linear function to match the endpoint values of the function. Of course, we can see from the graph that this rate of change calculation will vary depending on the interval selected.

We want to extend this idea. We want the rate of change over an interval of the form \([a,a]\). The rate of change at a single domain number.

We cannot use our \(\frac {rise}{run}\) idea for this extension, because \(run = 0\).

How would we go about translating out idea of rate of change over an interval to rate of change at a single domain number?

For instance, how would we think of the rate of change of \(Q\) at \(-3\)?

We could start by thinking of rates of change over small intervals near \(-3\).

\([-3, 4]\) would not be a good interval to use if you were interested in the rate-of-change near \(-3\), because \(4\) is too far away from \(-3\).

\([-3, -2.5]\) would probably give a better interval.

It seems like the smaller interval, the better.

The rate-of-change over the interval \([-3, -2.5]\) is

This is the slope of the secant line through the points \((-3, Q(-3))\) and \((-2.5, Q(-2.5))\).

That is looking more like a tangent line, than a secant.

Rate of Change Over a Single Number?

If we make a really small interval at \(-3\), then the slope of the secant line should give a good approximation of the tangent line. The slope of the secant line should approximate the slope of the tangent line.

The rate-of-change over the interval \([-3, -3+h]\) is

This is the slope of the secant line through the points \((-3, Q(-3))\) and \((-3+h, Q(-3+h))\), which is very near the tangent line at \((-3, Q(-3))\), when \(h\) is small.

By moving the value of \(h\) close to \(0\), we can get a good estimate of the slope of the tangent line.

The slope of the line tangent to the graph at \((-3, -6.4)\) looks to be about \(-0.8\).

Of course \(-3\) was nothing special. We can get slopes of tangent lines at any point on the graph. Let’s move around the graph and record the tangent line slopes.

The information we want to record is

(a)

the location of the tangent line, and

(b)

the slope of the tangent line

We could record the location of the tangent by giving the tangent point. A shorter way of recording the locaiton of the tangent line is to give the corersponding domain number.

For our tangent line above, we want to record \(-3\) and \(-0.8\).

We will record these as ordered pairs: (domain number, slope).

If we do this for a lot of points, then we create a collection of ordered pairs - just like a function.

We can represent the slope information \((-3, -0.8)\) as a plotted point.

The slope of the tangent line at \(-3\) is \(-0.8\), so we’ll plot the point \((-3,-0.8)\) to record this information. This will be a visual encoding of the tangent slope for the domain number \(-3\).

We could do this for a bunch of domain numbers.

(a)

select a domain number

(b)

select a very tiny interval beginning at that domain number

(c)

calculate the rate-of-change over the tiny interval

(d)

plot a point on the graph. First coordinate is the domain number. Second coordinate is the rate-of-change.

If we do this for every domain number, then we will have created a new function. Every domain number will be paired with the slope of the tangent line at the corresponding point.

We could graph this “slope function”.

This function, whose values are the rates-of-change of another function, is called the derivative of the other function, because it was derived from the original function.

\(\blacktriangleright \) Let \(f\) be a function. Then the derivative of \(f\) is denoted as \(f'\). People pronounce this as “f prime”.

Through the slope of the tangent line, we have invented a way of talking about the instantaneous rate of change of a function at a number, rather than over an interval.

The function \(Q(x) = \frac {(x+7)(x-5)}{5}\) is a quadratic and has a parabola for a graph. The vertex, \((-1, \tfrac {-36}{5})\), of the parabola has a horizontal tangent line, \(y=\tfrac {-36}{5}\).

The slope of this tangent line is \(0\). Therefore, \(Q'(-1) = 0\).

We could already figure this out, because \(Q\) is a quadratic function and the \(iRoC_Q\) is \(Q'\), the derivative of \(Q\).

Since \(Q\) is a quadratric and

\[ Q(x) = \frac {(x+7)(x-5)}{5} = \frac {1}{5} x^2 + \frac {2}{5} x - 7 \]

That gives us

\[ Q'(x) = \frac {2}{5} x + \frac {2}{5} \]

\[ Q'(1) = \frac {2}{5} (-1) + \frac {2}{5} = 0 \]

The Derivative

Graph of \(y = f(x) = e^{\tfrac {x}{5}}\)

The value of \(f'(4)\) is

positive negative