Unit 4 - Objective 2 - Concave Up and Concave Down

[Example 1] [Example 2] [Example 3] [Example 4] [Example 5]

 A function is concave up if the graph "holds water". The graph would be above its tangent lines. A function is concave down if the graph "spills water". The graph would be below its tangent lines. An inflection point is the point on the curve where the graph changes from concave up to concave down, or vice versa.

The sign of the second derivative indicates whether a function is concave up or concave down.

• If f ''(a) > 0, then y = f(x) is concave up when x = a.
• If f ''(a) < 0, then y = f(x) is concave down when x = a.

Determine whether the curve below is concave up or concave down at each point given.

y = f(x) = 3x³ - 6x² at x = -2 [or at the point (-2,-48)].
 Solution: f '(x) = 9x² - 12x f ''(x) = 18x - 12 f ''(-2) = -36 - 12 = -48 < 0 The second derivative is negative when x = -2 so the curve is concave down there.

[Example 1] [Example 2] [Example 3] [Example 4] [Example 5]

Use the second derivative as a test function for concave up and concave down. This tells us how to round out a graph when used with the increasing and decreasing information. Points where the second derivative is zero are possible inflection points and places where the graph could change concavity. Continuing the example on Worksheet 4-1:

Given y = f(x) = 2x³ - 3x² - 72x - 4 , find the intervals where f(x) is concave up and where it is concave down.
Steps Example
1.Find f ''(x)1. f '(x) = 6x² - 6x - 72
f ''(x) = 12x - 6

2.Find the possible inflection points,
where f ''(x) = 0
2.f ''(x) = 12x - 6 = 6(2x - 1) = 0

possible inflection point when x = 1/2
3.Put those x-values on the x-axis and 3.
4.Pick one test value in each interval
and put it in the 2nd derivative
to determine if it is positive or negative.
(I'm substituting into the factored
form of the 2nd derivative.)
4.

Test x = 0Test x = 1
f '' (0) = 6 (-1)f '' (1) = 6 (1)
f '' (0) = -6f '' (1) = +6
f '' (x) negativef '' (x) positive

 f (x) concave downfor x < 1/2 f (x) concave upfor x > 1/2

Note: From the concave up and down information above, we can tell that this function has an inflection point when x = 1/2.

The point of inflection is (1/2,-40.5) where the y value is f(1/2).

f(1/2) = 2(1/2)³ - 3(1/2)² - 72(1/2) - 4 = 1/4 -3/4 -36 - 4 = -40.5

[Example 1] [Example 2] [Example 3] [Example 4] [Example 5]

Combining the above information with the information obtained in Objective 4-1, we can now sketch a more accurate graph of this function. The only points that must be plotted are the ones corresponding to the critical values. For a more accurate graph, you may also plot the inflection points (or just estimate where they would be) and intercepts. Be sure to put numbers on your axes when you sketch a graph.

Example continued: Sketch y = f(x) = 2x³ - 3x² - 72x -4

xy
-3131required
4-212required
1/2-40.5optional

Using those points, sketch a graph which
increases to 131, decreases to -212,
then increases again and which is
concave down on the left of x = 1/2
and concave up on the right.This
sketch reperesents all of the graph,
so this function has no absolute
minimum and no absolute maximum.

x-axis scale = 1 unit
y-axis scale = 50 units

[Example 1] [Example 2] [Example 3] [Example 4] [Example 5]

Putting it all together: Sketch the graph of

Find the critical values and test y' inbetween to see if y is increasing or decreasing.

y' = 4x² - 12x = 4x(x² - 3) = 0

x = 0, x² - 3 = 0,

critical values: x = 0, ,

 x = -2 x = -1 x = 1 x = 2 y' = (-) y' = (+) y' = (-) y' = (+) y dec y inc y dec y inc

Find possible inflection points and test y'' in between to see if y is concave up or down.

y'' = 12x² - 12 = 12(x² - 1) = 12(x - 1)(x + 1) = 0

x = 1, x = -1, (possible inflection points)

 x = -2 x = 0 x = 2 y'' = (+) y'' = (-) y'' = (+) y c up y c dwn y c up

Plot critical point, then sketch, using information above. For plotting points, use decimal for

xy
-1.7-9
00
1.7-9
scale = 1 unit
on x and y axis

From this graph, we can say:

• There are two relative minimum points (negative square root of 3, -9) and (square root of 3, -9).
• There is a relative maximum at (0,0).
• There are inflection points at (-1,-5) and (1,-5).
• There is an absolute minimum of -9.
• There is no absolute maximum.

[Example 1] [Example 2] [Example 3] [Example 4] [Example 5]

You can sketch a graph even is you only have a description of how it behaves, rather than an equation.

Sketch the graph of a function, f(x), where

• f(0) = 0,
• f '(x) > 0 for all x,
• f ''(x) > 0 for x < 0, and
• f ''(x) < 0 for x > 0.

Solution:

• saying that f(0) = 0 gives one point on the graph, (0,0)
• f '(x) > 0 for all x means that the graph is increasing all the time (rising left to right)
• f ''(x) > 0 for x < 0 means the graph is concave up when x is negative. So round the curve to the left of the origin.
• f ''(x) < 0 for x > 0 means the graph is concave down when x is positive. So round the curve down to the right of the origin.
• The final graph might look something like this:
 Without more points, we don't know how steep it rises, but the shape should be about the same.

[Example 1] [Example 2] [Example 3] [Example 4] [Example 5]

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