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Curve Sketching and Relative Maxima and Minima the Goal

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Curve Sketching and Relative Maxima and Minima the Goal Curve Sketching and Relative Maxima and Minima The Goal The purpose of the first derivative sign diagram is to indicate where a function is increasing and decreasing. The purpose of this is to identify where the relative maxima and minima are, since we know that if a function switches from increasing to decreasing at some point and remains continuous there it has to have a relative maximum there (likewise for switching in the other direction for a relative minimum). This is known as the First Derivative Test. It is possible to come up with functions that are NOT continuous at a certain point and may not even switch direction but still have a relative maximum or minimum there, but we can't really deal with those here. Plot x - Floor x , x, 0, 4 , AspectRatio ® Automatic 1.0 0.8 @ @ D 8 < D 0.6 0.4 0.2 1 2 3 4 The purpose of the second derivative sign diagram is to fill in the shape of the function a little more for sketching purposes, since changes in curvature happen at different locations than changes in direction and are distinctive features of the curve. Addition- ally, in certain circumstances (when the second derivative exists and is not zero) the sign of the second derivative at a critical value will also indicate whether there is a maximum there (downward curvature, f''(x) < 0) or a minimum (upward curvature, f''(x) > 0). This is known as the Second Derivative Test. Also, the change in direction of the curvature also corresponds to the locations of the steepest slope, which turns out to be useful in certain applications. If you can identify the sign of the first and second derivative over the domain of the function it is possible to create a rough sketch by putting together some combination of four arcs. f' x < 0 and f'' x < 0 H L H L 2 CurveShape.nb f' x > 0 and f'' x < 0 H L H L f' x < 0 and f'' x > 0 H L H L f' x > 0 and f'' x > 0 H L H L Some definitions To indicate where a function is increasing or decreasing we need to identify points where it could possibly change from one to the other. Discontinuities - If there is a break in the function there is nothing to prevent it from sudden changes in value, slope, or even direction at that point. This is where the function doesn't exist, the limit of the function doesn't exist, or they are not the same. It could be a jump, a hole, or a vertical asymptote. Critical Values - If the function does exist and is continuous at some point it can switch from increasing to decreasing or vice versa by either having f'(x) = 0 or f'(x) not existing. Where either of these two conditions occur is called a critical value (CV for short). A function may also have a Horizontal Point of Inflection when f'(x) = 0 ( y = x3 at x = 0) or a Vertical Point of Inflection when f'(x) DNE (y = 3 x at x = 0) at a critical value. Neither of these involves a change of direction. CurveShape.nb 3 A function may also have a Horizontal Point of Inflection when f'(x) = 0 ( y = x3 at x = 0) or a Vertical Point of Inflection when f'(x) DNE (y = 3 x at x = 0) at a critical value. Neither of these involves a change of direction. Regular Inflection Point - One way to identify these points of inflection is that the curvature f''(x) switches sign there, which requires either the second derivative f''(x) to be zero or for it not to exist. Oddly, it IS possible for the curvature to switch at a minimum or maximum. To summarize - if there IS a change in direction it will definitely happen at either a discontinuity or a critical value. When the change in direction occurs at a critical value it will be either a relative maximum or minimum Procedure è Identify any discontinuities. This generally amounts to calculating where the denominator of f(x) is zero. There are some examples of functions that are discontinuous at other points but we would have to represent these with piecewise notation or something even more bizarre so we won't worry about that. è Calculate the derivative f'(x) and identify the CV. Typically this amounts to calculating the points where either the numerator or the denominator of f'(x) are zero but where the denominator of the original function f(x) is NOT zero. First Derivative Test è Determine the sign of the derivative between CV's and discontinuities. Pick a test point between, before, or after each CV or discontinuity and calculate f'(x) to see if it is positive or negative. è Diagram your results on the f'(x) sign diagram. è Classify any switch from + to - at a CV (but not at a discontinuity) as a relative maximum and any switch from - to + as a relative minimum Second Derivative Test è Calculate the second derivative f''(x) è Determine the sign of f''(x) at each CV è If f"(x) = 0 or does not exist at the CV then the test is inconclusive (might be a minimum or maximum but is likely a point of inflection). Otherwise if f''(x) > 0 it definitely indicates a minimum and if f''(x) < 0 it definitely indicates a maximum. For additional shape information è Identify the points of inflection when f"(x) = 0 or undefined. è Determine the sign of f"(x) between, before, or after any discontinuities and inflection points (you can use the values at the CV already calculated in the Second Derivative Test). è Diagram the resultes on an f''(x) sign diagram è Combine the first derivative and second derivative information on a sketch using the four arcs shown above. 4 CurveShape.nb Average Cost Example 1 238 900 avec x_ := 2900 x + x There is a vertical@ D asymptote at x = 0 but that is past the end of the domain, so we don't need to worry about any change in direction when x < 0. Find the derivative. Typically we need to simplify or factor it to determine where it is going to be zero or not exist avec' x 1 238 900 2900 -@ D x2 Factor avec' x 100 -12 389 + 29 x2 @ @ DD x2 I M The derivative does not exist at x = 0, but since it is not in the domain this does not qualify as a CV. Also any negative values of x are not of any interest. Solve for where the numerator is zero. cv = ToRules Reduce avec' x 0. && x > 0, x x ® 20.669 @ @ @ D DD Identify8 the sign of derivative< before and after this point (e.g. x = 10 and x = 30) Sign avec' 10 , Sign avec' 30 -1, 1 8 @ @ DD @ @ DD< Show sign8 of the< derivative, with shaded region being positive RegionPlot avec' x > 0, x, 0, 50 , y, 0, 1 , AspectRatio ® Automatic 1.0 0.00.20.40.60.8 0 @10 @ D 20 8 30 < 8 40 < 50 D Since it is decreasing before x = 20.669 and increasing after this is a minimum For analysis of the curvature, take the second derivative avec'' x 2 477 800 @ D x3 For x > 0 this is always positive, so there are no inflection points and the function always curves up. You can check the sign of the second derivative at the CV for a more explicit result of the Second Derivative Test Sign avec'' x . cv 1 @ @ D D As you can see, this is positive, indicating a minimum at x = 20.669. A graph of the sign of the second derivative shows that it is shaded (or positive) everywhere. CurveShape.nb 5 RegionPlot avec'' x > 0, x, 0, 50 , y, 0, 1 , AspectRatio ® Automatic 1.0 0.00.20.40.60.8 0 @10 @ D20 8 30 < 840 < 50 D We can show the two sign diagrams (f'(x) and f''(x)) together, along with some arcs to indicate the direction/shape of the curve 1.0 0.00.20.40.60.8 0 10 20 30 40 50 1.0 0.00.20.40.60.8 0 10 20 30 40 50 Here is the actual curve for comparison's sake Plot avec x , x, 0, 50 240 000 @ @ D 8 <D 220 000 200 000 180 000 160 000 140 000 10 20 30 40 50 Oxygen Purity Example 4 16 p t_ := 500 1 - + t + 4 t + 4 2 There is @a verticalD asymptote at t = -4 but that is past the end of the domain, so we don't need to worry about any change in H L direction at this point. Find the derivative. Typically we need to simplify or factor it to determine where it is going to be zero or not exist 6 CurveShape.nb Find the derivative. Typically we need to simplify or factor it to determine where it is going to be zero or not exist Simplify p' t 2000 -4 + t @ @ DD 4 + t 3 H L The derivative does not exist at t = -4, but since it is not in the domain this does not qualify as a CV. Also any negative values of H L t are not of any interest, practically speaking. Solve for where the numerator is zero. cv = ToRules Reduce p' t 0, t t ® 4 @ @ @ D DD Identify8 the sign< of derivative before and after this point (e.g.
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