Increasing and Decreasing Functions

A function f increases on a interval if $f(a) < f(b)$ whenever $a < b$ and a and b are points in the interval. This means that the graph goes up from left to right.

A function f decreases on a interval if $f(a) > f(b)$ whenever $a < b$ and a and b are points in the interval. This means that the graph goes down from left to right.

The Mean Value Theorem can be used to tell when a function increases and when it decreases.

If $f(x)$ is a differentiable function on an interval $a < x < b$ , then:

This makes sense, since the derivative gives the slope of the tangent line to the graph. Positive slope means the graph goes up from left to right and negative slope means the graph goes down from left to right.

In this way, I can use the derivative to obtain information about the shape of the graph. As an added benefit, I can tell whether a critical point is a local max or a local min.


Example. Find the intervals on which $y = 2x^3 + 3x^2 - 72x + 12$ increases and the intervals on which it decreases. Locate and classify any local extrema. Sketch the graph.

I'm going to find a little more information than the problem requested. Later on, I'll have a multi-step procedure for graphing a function; in this problem, I'll do the first few steps.

The domain is all real x.

The x-intercepts would probably be difficult to find, so I'll just continue. The y-intercept is $y = 12$ .

The derivatives are

$$y' = 6x^2 + 6x - 72 = 6(x + 4)(x - 3), \quad y'' = 1x + 6 = 6(2x + 1).$$

Notice that I want the derivatives in factored form. I won't use the second derivative $y''$ in these problems; you should get into the habit of finding it, however, since it will be part of the complete graphing procedure. ($y''$ will be needed to determine a curve's concavity.) $y'$ is defined for all x. $y' = 0$ for $x = -4$ and $x = 3$ . I set up a sign chart for $y'$ using the critical points as the break points. On each interval determined by the critical points, I pick a point at random and plug it into $y'$ .

If a point gives a positive value for $y'$ , then I know that $y'$ is positive on the interval, and hence that the function increases. I put a "+" above the interval and draw an upward-sloping line below it.

Likewise, if a point gives a negative value for $y'$ , then I know that $y'$ is negative on the interval, and hence that the function decreases. I put a "-" above the interval and draw an downward-sloping line below it.

$$\hbox{\epsfxsize=3in \epsffile{incdec1i.eps}}$$

Reading my sign chart, I see that the function increases for $x \le -4$ and for $x \ge 3$ . It decreases for $-4 \le x \le 3$ .

The upward and downward lines give a schematic picture of the graph of the function. Notice the shape of the graph at $x = -4$ : It shows that $x = -4$ is a local max.

Likewise, the shape of the schematic shows that $x = 3$ is a local min.

The use of $y'$ to classify a critical point as a max or a min is often called the First Derivative Test. By drawing a schematic picture with upward and downward lines, you remove the need to memorize the test: You can see from the schematic picture whether a point is a max or a min.

The schematic pictures tells me that the graph should have a general "up-down-up" shape. This is confirmed by the actual graph:

$$\hbox{\epsfysize=1.75in \epsffile{incdec1g.eps}}\quad\halmos$$


Example. Find the intervals on which $y = -\dfrac{9}{4}\dfrac{1}{x^4} +
   \dfrac{1}{2x^2}$ increases and the intervals on which it decreases. Locate and classify any local extrema. Sketch the graph.

The domain is $x \ne 0$ .

The x-intercepts are $x = \pm
   \dfrac{3}{\sqrt{2}}$ .

There is no y-intercept.

The derivatives are

$$y' = \dfrac{9}{x^5} - \dfrac{1}{x^3} = \dfrac{(3 - x)(3 + x)}{x^5}, \quad y'' = -\dfrac{45}{x^6} + \dfrac{3}{x^4} = \dfrac{3(x - \sqrt{15})(x + \sqrt{15})}{x^6}.$$

People often have trouble getting derivatives like these into the right form. The general procedure for derivatives with fractions or negative powers is:

If the derivative is a fraction, it will equal 0 when the top is 0 and it will be undefined when the bottom is 0. (There may be other undefined places if you have things like roots or logs, of course.)

In this case, $y'$ is undefined at $x = 0$ ; this is not a critical point (because y isn't defined at $x = 0$ ), but it counts as a break point on my sign chart. The break points on your sign chart include all points where $y' = 0$ or where $y'$ is undefined, regardless of whether the function is defined at those points. Finally, $y' = 0$ for $x = \pm 3$ .

$$\hbox{\epsfxsize=3in \epsffile{incdec2i.eps}}$$

The function increases for $x \le -3$ and for $0 < x \le 3$ . It decreases for $-3 \le x < 0$ and for $x \ge 3$ .

$x = -3$ is a local max and $x =
   3$ is a local min.

$$\hbox{\epsfysize=1.75in \epsffile{incdec2g.eps}}\quad\halmos$$


Example. Find the intervals on which $y = \dfrac{3}{7}x^{7/3} - 12x^{1/3}$ increases and the intervals on which it decreases. Locate and classify any local extrema. Sketch the graph.

The domain is all real x. (You can take the cube root of any number.)

The x-intercepts are $x = 0$ and $x = \pm
   \sqrt{28}$ .

The y-intercept is $y = 0$ .

The derivatives are

$$y' = x^{4/3} - 4x^{-2/3} = \dfrac{(x - 2)(x + 2)}{x^{2/3}}, \quad y'' = \dfrac{4}{3}x^{1/3} + \dfrac{8}{3}x^{-5/3} = \dfrac{4x^2 + 8}{3x^{5/3}}.$$

$y'$ is undefined at $x = 0$ ; since y is defined at $x = 0$ , this {\it is} a critical point. $y' =
   0$ for $x = \pm 2$ .

$$\hbox{\epsfxsize=3in \epsffile{incdec3i.eps}}$$

The function increases for $x \le -2$ and for $x \ge 2$ . It decreases for $-2 \le x \le 2$ .

$x = -2$ is a local max, $x = 2$ is a local min, and $x = 0$ is neither a max nor a min.

$$\hbox{\epsfysize=1.75in \epsffile{incdec3g.eps}}\quad\halmos$$

Example. Find the intervals on which $y = \dfrac{x^2 + 1}{x^2 - 1}$ increases and the intervals on which it decreases. Locate and classify any local extrema. Sketch the graph.

The domain is $x \ne \pm 1$ .

There are no x-intercepts. The y-intercept is $y = -1$ .

The derivatives are

$$y' = \dfrac{-4x}{(x^2 - 1)^2}, \quad y'' = \dfrac{4(3x^2 + 1)}{(x^2 - 1)^3}.$$

$y'$ is undefined for $x = \pm 1$ ; since y is also undefined for $x = \pm 1$ , these aren't critical points, though they are break points for my sign chart. $y' = 0$ for $x = 0$ .

$$\hbox{\epsfxsize=3in \epsffile{incdec4i.eps}}$$

The function increases for $x < -1$ and for $-1 < x \le 0$ . It decreases for $0 \le x < 1$ and for $x >
   1$ .

$x = 0$ is a local max.

$$\hbox{\epsfysize=1.75in \epsffile{incdec4g.eps}}\quad\halmos$$


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