Programmed tutorial: Derivative of the Absolute Value of a Quadratic Expression

Tutorial on the absolute value function using a linear expression and a parabola:

First, plot the graph of x² + 2x – 3 on paper. Then check your plot by clicking “Next”.

alt=" Graph of a quadratic Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!

Which values of x result in the y-coordinate being negative?

y is negative for x in – 3 < x < 1.

Indicate this interval on your graph, and then click “Next” to check.

alt="Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!

How does the graph of |x² + 2x – 3| differ from this graph?

The red portion is reflected above the x-axis.

Plot |x² + 2x – 3|, and then check by clicking “Next”.

alt=" Graph of the absolute value of a quadratic Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!

Now let’s discuss the algebraic sign of the slope (or the derivative) of this curve.
For what value of x is the slope zero?

We see that the tangent line is horizontal at x = – 1, so that is where the slope is 0.

Where is the slope positive?

For x > 1 and for – 3 < x < – 1.

Where is the slope negative?

For x < – 3 and for – 1 < x < 1.

Indicate these intervals with colors on your paper graph. Then click “Next” to check.

alt="Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!

For what colors on the x-axis is the slope positive?

For values of x in the blue or yellow intervals, the slope is positive.

For what colors on the x-axis is the slope negative?

For values of x in the red or green intervals, the slope is negative.

Will the derivative compare to the slope?

Yes, the derivative and the slope at every point are equal.

We haven’t mentioned the conditions at
x = – 3 and x = 1. What about that?

The derivative is undefined at those points.

Now we’ll relate these observations of the slope to the equation of the derivative:

Do we treat this with the same rule as for

No.

State the rule we must apply in terms of a general function u.

What effect does the

factor have?

It generates the correct algebraic sign.

What is the expression for u here?

u = x² + 2x – 3

What is the value of

Substitute these results in the rule for the derivative above.

Another way to write this result is

Will this always work out to a value of (2x + 2)?

No.

Which factor in

can cause the derivative to be – (2x + 2)?

Since the denominator is an absolute value, and is always positive, we need to look only at the x² + 2x - 3 factor in the numerator.

How do we find the values of x which make
x² + 2x – 3 negative?

We set x² + 2x – 3 < 0

Can this be factored?

Yes.

Do it.

We get (x + 3) * (x – 1) < 0

What are the conditions on these factors?

Either factor may be negative.

Express these conditions on each factor.

(x + 3) < 0, or (x – 1) < 0

What are the corresponding conditions on x?

x < – 3, or x < 1.

When we require that only one of the factors be negative, what are the conditions on x for that case?

– 3 < x < 1 will result in only one of the factors being negative. That in turn will make the quadratic negative, and from

we see that this condition results in a negative value of fractional part of the derivative. The following table summarizes the cases:

Expression	x < – 3	– 3 < x < – 1	– 1 < x < 1	x > 1
x² + 2x – 3	positive	negative	negative	positive
2x + 2	negative	negative	positive	positive
Product	negative	positive	negative	positive
Derivative	negative	positive	negative	positive
Color in diagram	red	yellow	green	blue

Does this table agree with the slopes in the figure above?

Yes.

The end. If you found this helpful and would recommend that I create more pages like this one, please let me know: Email to John Taylor