## The Calculus Primer (2011)

### Part VI. Further Applications of the Derivative

### Chapter 21. POINTS OF INFLECTION AND CURVE TRACING

**6—5.Points of Inflection.** When a curve changes its nature in such a way that there is a point at which the curve changes from concave downwards to concave upwards, or vice versa, that point is called a *point of inflection.*

From the discussion of §4—11, it will be seen, from the figure above, that at *R, f″*(*x*) is –, and at *S, f″*(*x*) is +; for as we move from left to right, or from *R* to *P*, the slope, *f*(*x*)*,* is decreasing, and when passing from *P* to *S,* the slope *f*′(*x*) is increasing. Thus, as *f*″(*x*) changes from − to + (or + to − ), it must pass through the value zero. Hence:

*at a point of inflection, f″*(*x*) *=* 0.[1]

Points of inflection may also be defined as points where

A general principle, given here without proof, is the following:

*Given a function f*(*x*) *and a specified point x*_{1} *such that f″*(*x*_{1}) = 0, *but f″*(*x _{1}*) ≠ 0,

*then x*

_{1}

*is the abscissa of a point of inflection on the curve y*=

*f*(

*x*).

It should be noted that at a point of inflection, the tangent to the curve passes *through* the curve. Points of inflection are very common. For example, algebraic polynomials of the form

*y =* *a*_{0}*x ^{n}* +

*a*

_{1}

*x*

^{n}^{−1}

*+ a*

_{2}

*x*

^{n-}^{2}

*+*· · · +

*a*

_{n}in general cannot have more than (*n* − 2) points of inflection. A parabola cannot have a point of inflection; for if *f*(*x*) = *ax*^{2} *+ bx* + *c*, then *f*′(*x*) = 2*ax* + *b*, and *f″*(*x*) *=* 2*a;* thus there is no value of *x* which makes *f″*(*x*) = 0. Similarly, a straight line can have no point of inflection, since if *f*(*x*) = *ax* + *b,* then *f′*(*x*) *= a*, *f*″(*x*) = 0, and *f″′*(*x*) = 0. The trigonometric functions, such as *y =* sin *x, y* = cos *x*, *y* = tan *x*, have an infinite number of points of inflection.

EXAMPLE 1.Determine the points of inflection of the curve whose equation is

*y* = *x*^{4} − 6*x*^{3} + 4*x*^{2} + 10*x* − 5,

*Solution.*

*f*(*x*) = *x*^{4} − 6*x*^{3} + 4*x*^{2} + 10*x* − 5,

*f′*(*x*) = 4*x*^{3} − 18*x*^{2} + 8*x* + 10,

*f″*(*x*) = 12*x*^{2} − 36*x* + 8,

*f″′*(*x*) = 24*x* − 36.

The values of *x* for which *f″*(*x*) = 0 are therefore found from the equation

12*x*^{2} − 36*x* + 8 = 0;

these values are

EXAMPLE 2.Test for points of inflection the curve of

*y* = *x*^{4} − 8*x*^{3} + 24*x*^{2} + 8*x*.

*Solution.*

*f′*(*x*) = 4*x*^{3} − 24*x*^{2} + 48*x* + 8,

*f″*(*x*) = 12*x*^{2} − 48*x* + 48,

*f″′*(*x*) = 24*x* − 48.

Values of *x* for which *f″*(*x*) = 0 are found from the equation

12*x*^{2} − 48*x* + 48 = 0,

which gives

(*x* − 2)^{2} = 0,or*x* = +2, +2.

However, for the value *x* = +2,

*f″′*(*x*) = 24(2) − 48 = 0;

hence there are no points of inflection, since *f″′*(*x*) = 0.

EXAMPLE 3.Find the points of inflection in the curve of *y* = cos *x.*

*Solution.*

*f′*(*x*) = − sin *x*,

*f″*(*x*) = − cos *x*,

*f″′*(*x*) *=* + sin *x*.

Solving *f″*(*x*) = 0 for *x*:

For these values of *x, f″′*(*x*) ≠ 0; hence points of inflection occur at

**EXERCISE 6—3**

*Test the following functions for points of inflection:*

**1.** *y* = *x*^{4} *+* 2*x*^{3} *−* 36*x*^{2} − *x*

**2.** *y* = *x*4 + 4*x*^{3} + 6*x*^{2} + 60*x*

**3.** *y* = *x*^{4} − 12*x*^{2}

**4.** *y =* 2*x*^{3}

**5.** *y = x*^{4}

**6.** *y* = *x5*

**7.** *y* = *e ^{x}*

**8.** *y* = *xe ^{x}*

**9.** *y* = *b* + (*x* − *a*)^{3}

**10.** *xy* − *x*^{3} = 1

**6—6.Curve Tracing.** When discussing curves and equations, an examination of the equation for *intercepts, extent, symmetry,* and *asymptotes* constitutes a useful and economical approach to determining the general shape of a curve. For the reader who may have forgotten the tests for symmetry with respect to the origin and the axes, we remind him:

(1) If an equation is unchanged by the substitution of *−x* for *x,* the curve is symmetrical with respect to the 7-axis.

(2) If an equation is unchanged by the substitution of −*y* for *y,* the curve is symmetrical with respect to the *X*-axis.

(3) If an equation is unchanged by the substitution of *−x* for *x* and −*y* for *y,* the curve is symmetrical with respect to the origin.

We also remind the reader who may have forgotten how to determine horizontal and vertical asymptotes:

(1) To find a vertical asymptote, solve the equation for *y*; if the solution is a fraction, set the denominator equal to zero and solve for *x.*

(2) To find a horizontal asymptote, solve the equation for *x*; if the solution is a fraction, set the denominator equal to zero and solve for *y*.

Limiting our discussion to single-valued functions, we may now add the determination of *maximum and minimum values* and *points of in-flection* as further tools for tracing a curve. When all these devices have been used, few additional points need to be calculated and plotted, and it will be possible to sketch the curve fairly accurately with a minimum of preliminary computation.

For convenience, the procedure for curve tracing is outlined below:

*Step* 1. Find the intercepts on the coordinate axes, if any.

*Step* 2. Test the curve for symmetry with respect to the axes and the origin.

*Step* 3. Determine the extent of the curve; that is, what values, if any, of either variable must be excluded.

*Step* 4. Find the asymptotes, if any, parallel to the coordinate axes; also, how *y* behaves for numerically (+ or − ) large values of *x*.

*Step* 5. Locate the critical points, where *f′*(*x*) = 0, and determine all maxima and minima.

*Step* 6. Locate the points at which *f″*(*x*) = 0, and determine all points of inflection.

EXAMPLE 1.Trace the curve of *y* = *x*^{3} − 3*x*^{2} *+ x +* 1.

*Solution.*

*f*(*x*) = *x*^{3} − 3*x*^{2} + *x +* 1,

*f′*(*x*) = 3*x*^{2} − 6*x* + 1,

*f″*(*x*) = 6*x* − 6,

*f″′*(*x*) = 6.

(1) Intercept on *Y*-axis = 1. Intercepts on *X*-axis are and 1, 1 + , and 1 – .

(2) Tests show that the curve is not symmetric to the origin or to either axis.

(3) Inspection shows that no values of *x* need be excluded.

(4) Inspection shows no asymptotes.

(5) Setting *f′(x)* equal to zero and solving gives critical values at *x* = Testing these further, it is found that a minimum exists at *x* = = 1.82; a maximum exists at *x* = = +.18.

(6) Setting *f″*(*x*) = 0 and solving, *x* = 1; since *f″′*(1) ≠ 0, there is a point of inflection at *x* = 1, i.e., the point (1,0). (That the abscissa of the point of inflection is also a root of *f*(*x*) = 0 is entirely accidental and of no significance.)

EXAMPLE 2.Trace the curve of

*Solution.*

(1)Intercept: *x* = 0, *y =* 0.

(2)No symmetry.

(3)Curve not defined when *x* = 1.

(4)Asymptotes: *x* − 1 = 0, and *y* − 1 = 0.

(5) no vaues of *x* for which *f′*(*x*) = 0; hence no maxima or minima.

(6)*f″*(*x*) = 2(*x* − 1)^{−3}; no values of *x* for which *f″*(*x*) = 0; hence no points of inflection.

EXAMPLE 3.Trace the curve:

*y* = 2e^{−x2}

*Solution.*

(1)When *x* = 0, *y* = 2. No value of *x* makes *y* = 0. Hence the only intercept is the point (0,2).

(2)Replacing *x* by −*x* shows curve symmetric with respect to *Y*-axis.

(3)No values of *x* need be excluded; all values of *x* yield values of *y* greater than zero but not greater than 2. Thus the curve lies entirely between the line *y* = 2 and the *X*-axis.

(4)As *x* becomes indefinitely large, positively or negatively, *y* approaches zero; thus the *X*-axis is an asymptote.

(5)Differentiating, *f′*(*x*) = −4*xe*^{−2}; solving *f′*(*x*) = 0 for *x,* we get *x* = 0 as the only critical value. As *x* passes through zero, *f′*(*x*) changes from + to −, and so (0,2) is a maximum point.

(6)Setting *f″*(*x*) = (16*x*^{2} − 4)e^{−x}_{2} = 0, we get *x* = ± Since *f″*(*x*) changes sign as *x* passes through either of these values, the points of inflection are (± ,2*e*^{–1/4}).

**EXERCISE 6—4**

*Trace, the following curves and draw the sketch in each case:*

**1.** *y* = *x*^{3} − 9*x*

**2.** *y* = *x*^{3} − *x*^{2} − 5*x* − 6

**5.** *y* = (*x −* 1)^{3}

**6.** *y* = *e*^{2x}

**7.** *y* = *e*^{−x2}

**8.** *y* = *x*(4 − *x*^{2})

**9.** *y* = *x*(log *x*)

**10.** *y* = *x* + sin *x*

**10.** Prove that (0,0) is not a point of inflection on the curve *y* = *x*^{6} − 2*x*^{4}.

**11.** Prove that no conic can have a point of inflection.