CONCAVITY AND INFLECTION POINTS
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- Philomena Bond
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1 CONCAVITY AND INFLECTION POINTS Find the Second Derivative of the function, f. Set the Second Derivative equal to zero and solve. Determine whether the Second Derivative is undefined for any x-values. To determine if these numbers are potential Inflection Points, make sure they are in the domain of the original function, f. If these numbers are NOT in the domain of the original function, f, and then stop here. Plot these numbers on a number line and test the regions with the Second Derivative. A positive result indicates the function is on that interval. A negative result indicates the function is Concave Down on that interval. The function has an Inflection Point at any value where the sign changes from positive to negative or negative to positive. Plug the x-value into the original function, f, to obtain the y-coordinate of the Inflection Point.
2 Example 1: Find all inflection points of the graph of f(x) = x 4 6x 2 + 8x +1. We know that f'(x) = 4x 3 12x + 8. We then proceed as follows: f''(x) = 12x x 2 12 = 12(x 2 1) = 12(x 1)(x + 1)= x=1, x = 1 defined for all x-values. Both 1 and 1 are in the domain of the original function, f(x) = x 4 6x 2 + 8x + 1 Plot these numbers on a number line and test the regions with the Second Derivative. Let s select a convenient number in the interval less than 1, between 1 and 1, and greater than 1. How about,, and 2, respectively? When we test in the Second Derivative, we obtain 12(-2) 2 12 = 36; when we test in the Second Derivative, we obtain 12() 2 12 = 12; and finally, when we test 2 in the Second Derivative, we obtain 12(2) 2 12 = 36. Therefore, the graph is concave up for x values less that 1 and greater than 1, and concave down between -1 and 1. Concave Down 1 1 The function has Inflection Points at -1 and 1 since the concavity changes. Plug these two values into the original function to obtain the y-coordinates of the Inflection Points: f( 1)=( 1) 4 6( 1) 2 + 8( 1) + 1 = 3 f(1)=(1) 4 6(1) 2 + 8(1) + 1 = 13 So, ( 1, 3) and (1, 13) are Inflection Points.
3 Example 2: Find all inflection points of the graph of f(x) = x 4. We know that f'(x) = 4x 3. We then proceed as follows: f''(x) = 12x 2 12x 2 = X = defined for all x-values. is in the domain of the original function, f(x) = x 4. Plot this number on a number line and test the regions with the Second Derivative. Let s select a convenient number in the interval less than zero. How about -1? Then we select a convenient number in the interval greater than zero, How about 1? Now, when we test -1 and 1 in the Second Derivative, we obtain 12(-1) 2 = 12, and 12(1) 2 = 12 Since both results are positive, the graph is concave up on both intervals No change in concavity. Consequently, (, ) is NOT an Inflection Point.
4 Example 3: Find all inflection points of the graph of f(x) = x We know that f'(x) = 1 x 3. We then proceed as follows: 3 f''(x) = 9 x 5 3 = 9x 5 3 never. undefined when x =. is in the domain of the original function, f(x) = x Plot this number on a number line and test the regions with the Second Derivative. Let s select a convenient number in the interval less than zero. How about -1? Then we select a convenient number in the interval greater than zero, How about 1? Now, when we test -1 and 1 in the Second Derivative, we obtain = = 2 9( 1) and 9(1) Since the first result is positive and the second is negative, the graph is concave up for all values less than and concave down for all values greater than. Concave Down The function has an Inflection Point at since the concavity changes. Plug into the original function to obtain the y-coordinates of the Inflection Point: f() = () = 2 So, (, 2) is an Inflection Point.
5 Example 4: Find all inflection points of the graph of f(x) = 1 x. We know that f'(x) = -1x -2. We then proceed as follows: f''(x) = 2x -3 = 2 x 3 never. undefined when x =. is NOT in the domain of the original function, f(x) = 1 x. Therefore, there are NO INFLECTION POINTS due to the fact that the original function, f, is not defined at. Let s say you did perform the number line test: 2 If we test -1 and 1 in the Second Derivative, we obtain = ( 1) and 2 = 2 3 (1) 3 Concave Down It is very tempting to conclude that since the graph is concave down for all x values less than, and concave up for all x values greater than, then must be a point of inflection. However, the fact that a change in concavity occurs is not, of itself, a guarantee that there is an Inflection Point. You must make sure that the function is defined at the number, and in this problem, it was not. Therefore, there is no Inflection Point.
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