# Elements of the Differential and Integral Calculus/Chapter VIII part 3

 Elements of the Differential and Integral Calculus by William Anthony Granville Chapter VIII, § 85–86

85. Points of inflection. Definition. Points of inflection separate arcs concave upwards from arcs concave downwards.[1] Thus, if a curve $y = f(x)$ changes (as at B) from concave upwards (as at A) to concave downwards (as at C), or the reverse, then such a point as B is called a point of inflection.

From the discussion of §84 it follows at once that at A, $f''(x) = +$, and at C, $f''(x) = -$. In order to change sign it must pass through the value zero;[2] hence we have

(23) at points of inflection, $f''(x) = 0$.

Solving the equation resulting from (23) gives the abscissas of the points of inflection. To determine the direction of curving or direction of bending in the vicinity of a point of inflection, test $f''(x)$ for values of x, first a trifle less and then a trifle greater than the abscissa at that point.

If $f''(x)$ changes sign, we have a point of inflection, and the signs obtained determine if the curve is concave upwards or concave downwards in the neighborhood of each point of inflection.

The student should observe that near a point where the curve is concave upwards (as at A) the curve lies above the tangent, and at a point where the curve is concave downwards (as at C) the curve lies below the tangent. At a point of inflection (as at B) the tangent evidently crosses the curve.

Following is a rule for finding points of inflection of the curve whose equation is $y = f(x)$. This rule includes also directions for examining the direction of curvature of the curve in the neighborhood of each point of inflection.

FIRST STEP. Find $f''(x)$.

SECOND STEP. Set $f''(x) = 0$, and solve the resulting equation for real roots.

THIRD STEP. Write $f''(x)$ in factor form.[/itex]

FOURTH STEP. Test $f''(x)$ for values of x, first a trifle less and then a trifle greater than each root found in the second step. If $f''(x)$ changes sign, we have a point of inflection.

When $f''(x) = +$, the curve is concave upwards.[3]

When $f''(x) = -$, the curve is concave downwards.

EXAMPLES

Examine the following curves for points of inflection and direction of bending.

1. $y = 3x^4 - 4x^3 + 1$.

 Solution. $f(x)$ $= 3x^4 - 4x^3 + 1$. First step. $f''(x)$ $= 36x^2 - 24x$. Second step. $36x^2 - 24x$ $= 0$ ∴ $x = \frac{2}{3}$ and $x = 0$, critical values. Third step. $f''(x)$ $= 36x(x - \frac{2}{3})$. Fourth step. When $x < 0, f''(x) = +$; and when $x > 0, f''(x) = -$. ∴ curve is concave upwards to the left and concave downwards to the right of $x = 0$ (A in figure). When $x < \frac{2}{3}, f''(x) = -$; and when $x > \frac{2}{3}, f''(x) = +$. ∴ curve is concave downwards to the left and concave upwards to the right of $x = \frac{2}{3}$ (B in figure).

The curve is evidently concave upwards everywhere to the left of A, concave downwards between A $(0, 1)$ and B $(\frac{2}{3}, \frac{11}{27})$, and concave upwards everywhere to the right of B.

2. $(y - 2)^3 = (x - 4)$,

 Solution. $y$ $= 2 + (x - 4)^{-\frac{1}{3}}$. First step. $\frac{dy}{dx}$ $= \frac{1}{3}(x - 4)^{-frac{2}{3}}$, Second step. When x = 4, both first and second derivatives are infinite. Third step. When $x < 4, \frac{d^2 y}{dx^2} = +$; but when $x > 4, \frac{d^2 y}{dx^2} = -$.

We may therefore conclude that the tangent at (4, 2) is perpendicular to the axis of X, that to the left of (4,2) the curve is concave upwards, and to the right of (4, 2) it is concave downwards. Therefore (4, 2) must be considered a point of inflection.

 3. $y = x^2$. Ans. Concave upwards everywhere. 4. $y = 5 - 2x - x^2$. Concave downwards everywhere. 5. $y = x^3$. Concave downwards to the left and concave upwards to the right of (0, 0). 6. $y = x^3 - 3x^2 - 9x + 9$. Concave downwards to the left and concave upwards to the right of (1, -2). 7. $y = a + (x - b)^3$. Concave downwards to the left and concave upwards to the right of (b, a). 8. $a^2 y = \frac{x^3}{3} - a x^2 + 2 a^3$. Concave downwards to the left and concave upwards to the right of $(a, \frac{4a}{3})$. 9. $y = x^4$. Concave upwards everywhere. 10. $y = x^4 - 12x^3 + 48x^2 - 50$. Concave upwards to the left of x = 2, concave downwards between x = 2 and x = 4, concave upwards to the right of x = 4. 11. $y = \sin x$. Points of inflection are $x = n\pi$, n being any integer. 12. $y = \tan x$. Ans. Points of inflection are $x = n\pi$, n being any integer.

13. Show that no conic section can have a point of inflection.

14. Show that the graphs of $e^x$ and $\log x$ have no points of inflection.

86. Curve tracing. The elementary method of tracing (or plotting) a curve whose equation is given in rectangular coordinates, and one with which the student is already familiar, is to solve its equation for y (or x), assume arbitrary values of x (or y), calculate the corresponding values of y (or x), plot the respective points, and draw a smooth curve through them, the result being an approximation to the required curve. This process is laborious at best, and in case the equation of the curve is of a degree higher than the second, the solved form of such an equation may be unsuitable for the purpose of computation, or else it may fail altogether, since it is not always possible to solve the equation for y or x.

The general form of a curve is usually all that is desired, and the Calculus furnishes us with powerful methods for determining the shape of a curve with very little computation.

The first derivative gives us the slope of the curve at any point; the second derivative determines the intervals within which the curve is concave upward or concave downward, and the points of inflection separate these intervals; the maximum points are the high points and the minimum points are the low points on the curve. As a guide in his work the student may follow the

Rule for tracing curves. Rectangular coördinates.

FIRST STEP. Find the first derivative; place it equal to zero; solving gives the abscissas of maximum and minimum points.

SECOND STEP. Find the second derivative; place it equal to zero; solving gives the abscissas of the points of inflection.

THIRD STEP. Calculate the corresponding ordinates of the points whose abscissas were found in the first two steps. Calculate as many more points as may be necessary to give a good idea of the shape of the curve. Fill out a table such as is shown in the example worked out.

FOURTH STEP. Plot the points determined and sketch in the curve to correspond with the results shown in the table.

If the calculated values of the ordinates are large, it is best to reduce the scale on the Y-axis so that the general behavior of the curve will be shown within the limits of the paper used. Coordinate plotting paper should be employed.

EXAMPLES

Trace the following curves, making use of the above rule. Also find the equations of the tangent and normal at each point of inflection.

1. $y = x^3 - 9x^2 + 24x - 7$.

Solution. Use the above rule.

 First step. $y'$ $= 3x^2 - 18x + 24$, $3x^2 - 18x + 24$ $= 0$ x $= 2, 4$. Second step. $y''$ $= 6x - 18$, $6x - 18$ $= 0$, $x$ $= 3$,

Third step.

x y y' y'' Remarks Direction of Curve
0 -7 + - max.
pt. of infl.
min.
concave down
2 13 0 -
3 11 - 0 concave up
4 9 0 +
6 29 + +

Fourth step. Plotting the points and sketching in the curve, we get the figure shown. To find the equations of the tangent and normal to the curve at the point of inflection $P_1 (3, 11)$, use formulas (1), (2), pp. 76, 77 [§65]. This gives $3x + y = 20$ for the tangent and $3y - x = 30$ for the normal.

2. $y = x^3 - 6x^2 - 36x + 5$.

Ans. Max. (-2, 45); min. (6, -211); pt. of infl. (2, -83); tan. $y + 48x - 13 = 0$; nor. $48y - x + 3986 = 0$.

3. $y = x^4 - 2x^2 + 10$.

Ans. Max. (0, 10); min. (± 1, 9); pt. of infl. $\left ( \pm \frac{1}{\sqrt{3}}, \frac{85}{9} \right )$.

4. $y = \frac{1}{2}x^4 - 3x^2 + 2$.

Ans. Max. (0, 2); min. $(\pm \sqrt{3}, -\frac{5}{2} )$; pt. of infl.$(\pm 1, -\frac{1}{2})$.

5. $y = \frac{6x}{1 + x^2}$.

Ans. Max. (1, 3); min. (-1, -3); pt. of infl. (0, 0), $\left ( \pm \sqrt{3}, \pm \frac{3\sqrt{3}}{2} \right )$.

6. $y = 12x - x^3$.

Ans. Max. (2, 16); min. (-2, -16); pt. of infl. (0, 0).

7. $4y + x^3 - 3x^2 + 4 = 0$.

Max. (2,0); min. (0, -1).

 8. $y = x^3 - 3x^2 - 9x + 9$. 20. $y = 3x - x^3$. 9. $2y + x^3 - 9x + 6 = 0$. 21. $y = x^3 - 9x^2 + 15x - 3$. 10. $y = x^3 - 6x^2 - 15x + 2$. 22. $x^2 y = 4 + x$. 11. $y(1 + x^2) = x$. 23. $4y = x^4 - 6x^2 + 5$. 12. $y = \frac{8a^3}{x^2 + 4a^2}$. 24. $y = \frac{x^3}{x^2 + 3a^2}$. 13. $y = e^{-x^2}$. 25. $y = \sin x + \frac{x}{2}$. 14. $y = \frac{4 + x}{x^2}$. 26. $y = \frac{x^2 + 4}{x}$. 15. $y = (x + l)^{\frac{2}{3}}(x - 5)^2$. 27. $y = 5x - 2x^2 - \frac{1}{3}x^3$. 16. $y = \frac{x + 2}{x^3}$. 28. $y = \frac{1 + x^2}{2x}$. 17. $y = x^3 - 3x^2 - 24x$. 29. $y = x - 2\sin x$. 18. $y = 18 + 36x - 3x^2 - 2x^3$. 30. $y = \log \cos x$. 19. $y = x - 2 \cos x$. 31. $y = \log (1 + x^2)$.

1. Points of inflection may also be defined as points where
 (a) $\frac{d^2 y}{dx^2} = 0$ and $\frac{d^2 y}{dx^2}$ changes sign, or (b) $\frac{d^2 x}{dy^2} = 0$ and $\frac{d^2 x}{dy^2}$ changes sign.
2. It is assumed that $f'(x)$ and $f''(x)$ are continuous. The solution of Ex. 2, p. 127 [§85], shows how to discuss a case where $f'(x)$ and $f''(x)$ are both infinite. Evidently salient points (see p. 258 [§156]) are excluded, since at such points $f'(x)$ is discontinuous.
3. This may be easily remembered if we say that a vessel shaped like the curve where it is concave upwards will hold (+) water, and where it is concave downwards will spill (-) water.