CURVE SKETCHING

The First Derivative Test

The first derivative of a function represents the slope of the tangent to that function's curve at each point. (see The Derivative: or What's the Slope of Your Surfboard? in MathRoom Musings) So it's pretty obvious that points where the tangent or surfboard slope = 0 must be points at the crest or trough of the wave since a line with slope = 0 is a horizontal line.

The horizontal line which touches the curve at x = 4 has slope = 0. Notice that it's at the vertex, minimum or critical point. Every tangent left of x = 4 will have a negative slope and every tangent right of x = 4 will have a positive slope. Note that left of x = 4, f(x) is decreasing and right of x = 4, f(x) is increasing -- so the sign of the 1st derivative determines where f(x) increases and decreases.

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 The First Derivative Test first derivative info function curve info points where f '(x) = 0 or º critical points (max, min, point of inflection) on intervals where f '(x) > 0 f(x) is increasing on intervals where f '(x) < 0 f(x) is decreasing if over an interval f '(x) goes + 0 - then local max at point where f '(x) = 0 if over an interval f '(x) goes - 0 + then local min at point where f '(x) = 0

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 first derivative test second derivative test graph properties list asymptotes sample graph holes in the graph practice solutions

The Second Derivative Test

Since the sign of the 2nd derivative indicates where the slope of the tangent {f '(x)} to the function f(x) is increasing and decreasing, it determines the shape of the curve for f(x).

Where f ''(x) > 0 , f '(x) is increasing -- so the slopes of the tangents to f(x) are increasing and vice versa.

Let's make a sketch of a bunch of tangents with increasing and decreasing slopes to see what shapes they define.

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It should be obvious then that if there is a point where f ''(x) = 0, the curve is changing convexity at that point.

This is called a point of inflection.

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 The Second Derivative Test second derivative info function curve info points where f ''(x) = 0 point of inflection on intervals where f ''(x) > 0 f(x) is concave up on intervals where f ''(x) < 0 f(x) is concave down if f '(c) = 0 and f "(x) < 0 , then local max at (c f(c) ) if f '(c) = 0 and f "(x) > 0 , then local min at (c f(c) )

if f '(c) = 0 and f '' (c) > 0, then (c, f(c) ) is a local minimum
since it's a critical point on a concave up curve.

if f '(c) = 0 and f '' (c) < 0, then (c, f(c) ) is a local maximum
since it's a critical point on a concave down curve.

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 first derivative test second derivative test graph properties list asymptotes sample graph holes in the graph practice solutions

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Graph Properties List:

When we're asked to sketch the graph of a given function, we use the 1st and 2nd derivative tests and other methods to make a list of properties for the function.

Thanks to an old Calculus text book written by a fellow named Bonic, we have an acronym for the first letters of the items in the list. It is M2I2ACIDS an acronym for:

 M2 = max, min, I2 = intercepts, inflection points, A = asymptotes, C = convexity, I = increasing D = decreasing S = symmetry

Below is such a list. Note the numbers beside the list items. They tell us in what order to "fill in the blanks".

5) MAXIMUM: f '(x) = 0, f '' (x) < 0

6) MINIMUM: f '(x) = 0, f '' (x) > 0

1) INTERCEPTS: set x = 0 for y-int, set y = 0 for x-int.

9) INFLECTION POINT: f '' (x) = 0

2) ASYMPTOTES: vert: x's that make only the denominator = 0

horiz: limit as x approaches º for f(x).
oblique: see note below.

7) CONCAVE UP: f '' (x) > 0.

8) CONCAVE DOWN: f '' (x) < 0.

3) INCREASING: f '(x) > 0.

4) DECREASING: f '(x) < 0.

10)SYMMETRY: look at the graph.

We do the analysis in 3 steps.

2) find the 1st derivative fill in the info, and
3) find the 2nd derivative, fill in the info. Then graph the curve.

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 first derivative test second derivative test graph properties list asymptotes sample graph holes in the graph practice solutions

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Asymptotes

Remember that asymptotes are lines and therefore we must write an equation for the asymptote line. We cannot just say the asymptote is 3 or -2. It must be x = 3 or x = -2 !

VERTICAL ASYMPTOTES are caused by x-values that make only the denominator = 0

The vertical asymptotes of are x = 2 and x = - 4.

X-values that make both numerator and denominator = 0 cause a hole in the graph. In such cases, there is only a single point missing rather than an infinite set of points. Such a hole is called a removeable jump or discontinuity since defining a value for f(x) at that point eliminates the discontinuity. See note Holes in the Graph.

HORIZONTAL ASYMPTOTES define the behavior of the curve at the extremes of the domain -- that is -- what's happening to the y-values as x approaches negative or positive infinity.

This is why we find to determine the horizontal asymptotes. The graph may cross the y-value of the asymptote at certain points -- but will only approach this value as x t º.

Finding limits at infinity:

If f(x) is a rational function (a fraction), and the degree of the numerator is less than or equal to the degree of the denominator, divide all terms by the highest power of x and then set x = º. Anything with a denomintor of º becomes 0.

If the degree of the numerator is greater than the degree of the denominator, divide the fraction to get a polynomial and perhaps a fraction remainder. Now we're looking at oblique asymptotes.

OBLIQUE ASYMPTOTES:

If y = and the degree of f(x) is > the degree of g(x), divide g(x) into f(x) to obtain
y = ax + b + . The line y = ax + b is the oblique asymptote.

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 first derivative test second derivative test graph properties list asymptotes sample graph holes in the graph practice solutions

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Sample Graph:

We will list the M2I2ACIDS features of and graph it.

1) INTERCEPTS: if x = 0 then y = 0 so (0, 0) is both intercepts.

2) ASYMPTOTES: Vertical: if x = 4 then denom. = 0 so x = 4 is a vertical asymptote.

Horizontal: = 3, so y = 3 is the horizontal asymptote.

first derivative:

f '(x) = so f '(x) < 0 for all x since the numerator is a negative constant and the denominator is a perfect square. ( > 0 )

3) INCREASING: none since f '(x) < 0 always

4) DECREASING: for all x

5) MAXIMUM: none since f '(x) ! 0.

6) MINIMUM: none since f '(x) ! 0.

second derivative:

f '' (x) = f '' (x) ! 0.
f '' (x) > 0 when x > 4, and f ' ' (x) < 0 when x < 4.

7) CONCAVE UP: x > 4

8) CONCAVE DOWN: x < 4

9) INFLECTION POINT: none since f '' (x) is never = 0.

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graph

5) MAXIMUM: none

6) MINIMUM: none

1) INTERCEPTS: (0,0)

9) INFLECTION POINT: none

2) ASYMPTOTES: vert: x = 4 horiz: y = 3

7) CONCAVE UP: x > 4

8) CONCAVE DOWN: x < 4

3) INCREASING: never

4) DECREASING: for all x

10)SYMMETRY: to the point (4, 3).

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 first derivative test second derivative test graph properties list asymptotes sample graph holes in the graph practice solutions

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HOLES IN THE GRAPH occur when an x value (x = a) makes both numerator and denominator of a fraction = 0. To find the missing y value, find the .

ex: f(x) = has a hole at x = 3. Proceed like this:

. There is a hole in the graph at (3, 1/6).

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Practice

1) Apply the 1st and 2nd derivative tests to find the critical and inflection points, increasing, decreasing, concave up and concave down intervals, for each function. Label each critical point a max or min.

 a) f(x) = x 3 - 3x 2 + 1 b) g(x) = 3x 4 - 4x 3

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2) Draw the graph of and list the M2I2ACIDS features (& any other important features) for

*Clearly indicate f '(x), f ' '(x) and how you found the asymptotes.(hint: factor the demominator).

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3) Sketch the graph of the continuous function f(x) which satisfies all of these conditions:

 f(0) = 4 f(2) = 2 f(5) = 6 f '(0) = f '(2) = 0 f '(x) > 0 if | x - 1 | > 1 f '(x) < 0 if | x - 1 | < 1 f ''(x) < 0 if x < 1 if | x - 4 | < 1 f ''(x) > 0 if | x - 2 | < 1 if x > 5

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 first derivative test second derivative test graph properties list asymptotes sample graph holes in the graph practice solutions

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Solutions

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 a) f(x) = x 3 - 3x 2 + 1f ' (x) = 3x 2 - 6x = 3x (x - 2)f ' (x) = 0 at x = 0 and x = 2 (critical pts.)f ' (x) < 0 on 0 < x < 2, decreasingf ' (x) > 0 on x < 0 and x > 2, increasingf "(x) = 6x - 6 = 6(x - 1)f "(x) = 0 at x = 1, point of inflectionf "(x) < 0 on x < 1, concave downf "(x) > 0 on x > 1, concave upx = 0 is a max (on concave down interval)x = 2 is a min (on concave up interval). b) g(x) = 3x 4 - 4x 3 g '(x) = 12x 3 - 12x 2 = 12x 2 (x - 1)g '(x) = 0 at x = 0 and x = 1 (critical pts.)g '(x) < 0 on x < 1, decreasingg '(x) > 0 on x > 1, increasingg "(x) = 36x 2 - 24x = 12x (3x - 2)g "(x) = 0 at x = 0 and x = 2/3, inflection pts.g "(x) < 0 on 0 < x < 2/3, concave downg "(x) > 0 on x < 0 and x > 2/3, concave upx = 0 is a point of inflectionx = 1 is a minimum (concave up interval)this is a parabola opening upwards.

This function has 2 vertical asymptotes at x = -1 and x = 1

h '(x) = 0 at x = 0, (critical point)

h '(x) < 0 when x > 0, since the denominator is always > 0.(decreasing)

h '(x) > 0 when x < 0, (increasing)

h "(x) ! 0 since the minimum value of the numerator is = 1, no inflection points.

h "(x) < 0 on -1 < x < 1, concave down

h "(x) > 0 on x < -1 and x > 1, concave up

x = 0, critical point is a maximum (on concave down interval).

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2) Draw the graph of and list the M2I2ACIDS features for

This function has a hole in the graph at x = 3 since both numerator and denominator = 0.

We do the analysis on , then find the limit as x t 3 for the original function.

Step 1: intercepts & asymptotes

When x = 0, g(0) = -1/2 -- so (0, -1/2) is y-intercept

VA: x = 2, HA: y = 0

Step 2: first derivative

g '(x) < 0 ¼ x since the numerator is a negative constant, denominator is a perfect square.

g(x) is decreasing ¼ x

Since g '(x) ! 0, no max, no min.

Step 3: second derivative

g "(x) ! 0, no points of inflection.

g "(x) < 0 when x < 2, concave down.

g "(x) > 0 when x > 2, concave up.

Hole in the graph at x = 3

, hole at (3, 1) -- only 1 hole -- so no symmetry.

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3) Sketch the graph of the continuous function f(x) which satisfies all of these conditions:

 f(0) = 4so (0, 4) is a point f(2) = 2so (2, 2) is a point f(5) = 6so (5, 6) is a point f '(0) = f '(2) = 0critical points f '(x) > 0 if | x - 1 | > 1increasing intervalson x < 0, and x > 2 f '(x) < 0 if | x - 1 | < 1decreasing intervalson 0 < x < 2 f ''(x) < 0 concave down if x < 1 if | x - 4 | < 13 < x < 5 f ''(x) > 0 concave up if | x - 2 | < 1on 1 < x < 3 if x > 5

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 first derivative test second derivative test graph properties list asymptotes sample graph holes in the graph practice solutions

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