Curve Tracing, Tangent and Normal Lines

Now we shall see how some of the things we have learned can help us to get a quick idea of the appearance of the graph of a function $f(x)$. In practice, we are usually concerned with functions which have as many derivatives as we wish (the second derivative f''(x) is simply the derivative of the derivative f'(x), and so on to higher derivatives) except at a few points whose exceptional behavior is fairly clear from the start.

To begin with, we must extend our definition of maximum and minimum to include relative maxima and minima. Let $f(x)$ be defined in the interval $a\le x\le b$. A point , is called a relative maximum of $f(x)$ if there is a neighborhood of $x_0$, that is, an interval $c\le x\le d$ such that and , with the property that $x_0$ is a maximum for $f(x)$ in this interval. Relative minimum is similarly defined. If $f(x)$ is continuous in $a\le x\le b$ and differentiable for , the same properties hold in any sub-interval $c\le x\le d$; our earlier considerations of maxima and minima then tell us that \boldsymbol{f'(x_0) = 0} whenever $x_0$ is a relative maximum or a relative minimum.

The converse is false, as is easily seen by considering the curve $y=x^3$ at $x_0=0$. Here y' = 3x^2, so y'(0) = 0, but 0 is neither a relative maximum nor a relative minimum, as is evident from the graph.

Therefore once we have determined the points where y' = 0, we must determine whether they are relative maxima, relative minima, or neither of these.

Suppose f'(x_0) = 0, and suppose we also calculate f''(x_0) and find f''(x_0) > 0. Then for small $h$,

\frac{f'(x_0+h) - f'(x_0)}{h} = \frac{f'(x_0+h)}{h} > 0 .

Thus if h > 0, f'(x_0+h) > 0; if h < 0, f'(x_0+h) < 0. Now consider an interval $x_0-\delta \le x\le x_0+\delta$ small enough so that for all $h$ such that 0 < h \le \delta, f'(x_0+h) > 0 and for all $h$ such that -\delta \le h < 0, f'(x_0+h) < 0. Then $x_0$ **is a minimum for $f(x)$ in this interval.** For if $x_1$ is any point of the interval and , then f(x_1) - f(x_0) = f'(\xi)(x_1 - x_0), where or $\xi =x_0+(x_1-x_0)$, with . Thus f'(\xi) > 0, and . In similar fashion we show that if $x_2$ is a point of the smaller interval and , then . This proves the assertion, and shows in addition that $x_0$ is the only minimum of $f(x)$ in $x_0-\delta \le x\le x_0+\delta$. We have now proved the following:

By the same type of reasoning, we can prove:

If f'(x_0) = 0 = f''(x_0), our information is still incomplete; we could have behavior like that of $y=x^3$ at 0, or a minimum as in the case of $y=x^4$ at 0, or a maximum as in the case of $y=-x^4$ at 0. We could even have more complicated behavior; consider the following function at $x=0$:

f(x) = \left\{ \begin{aligned} x^4 \sin \frac{1}{x}, \quad x \neq 0 \\ 0 \quad \text{if } x = 0 \end{aligned} \right\}

is really impossible to sketch with any accuracy in a neighborhood of 0 because the breadth of the curves we actually draw will not allow us the space to represent very many of its fluctuations.

Now let us look a little more closely at f''(x). Suppose f''(x_1) > 0 at some point $x_1$. Then for small $h$, the difference quotient

\frac{f'(x_1+h) - f'(x_1)}{h}

is positive; if h > 0, f'(x_1+h) > 0, and if h < 0, f'(x_1+h) < 0. But f'(x_1) is the slope of the tangent to the curve at $x_1$, so this tangent is given by the equation y = f'(x_1)x + b, where $b$ is yet to be determined. But we must have $(x_1,f(x_1))$ on this line, or f(x_1) = x_1 f'(x_1) + b, and b = f(x_1) - x_1 f'(x_1). Thus the tangent line at ${𝒙}_{\mathbf{1}}$ is

y = x f'(x_1) + (f(x_1) - x_1 f'(x_1)) .

Now let $x$ be in a neighborhood of $x_1$ so small that whenever $x_1+h$ is in this neighborhood,

\frac{f'(x_1+h) - f'(x_1)}{h} > 0 .

Then f(x) - f(x_1) = f'(\xi)(x - x_1), where $\xi$ is between $x$ and $x_1$, therefore is surely in this neighborhood of $x_1$. If , then , and f'(\xi) > f'(x_1) as we have seen above. Therefore

f(x) - f(x_1) > f'(x_1)(x - x_1),

or

f(x) > x f'(x_1) + (f(x_1) - x_1 f'(x_1)) .

If , then f'(\xi) < f'(x_1), and

f(x_1) - f(x) < f'(x_1)(x_1 - x),

or

f(x) > x f'(x_1) + (f(x_1) - x_1 f'(x_1)) .

In either case, the value of the original function is greater than the value of the function describing the tangent line, or the curve lies above its tangent at ${𝒙}_{\mathbf{1}}$ (meaning, of course, in a neighborhood of $x_1$). In this case we say that the curve is concave upward at ${𝒙}_{\mathbf{1}}$. Likewise, if f''(x_1) < 0, the curve lies below its tangent at $x_1$, in a neighborhood of $x_1$, and we call it concave downward at $x_1$. Thus the second derivative is related to the curvature of the curve; for example, if f''(x_1) is large and positive, the curve bends sharply upward at $x_1$; if f''(x_1) is large and negative, the curve bends sharply downward at $x_1$.

Suppose now that the graph of $f(x)$ crosses its tangent at a point $x_0$. Then $x_0$ is called a point of inflection of $f(x)$. We have seen that if f''(x_0) > 0, the curve lies above its tangent in a neighborhood of $x_0$, and that if f''(x_0) < 0, it lies below the tangent. Therefore the only possible candidate for a point of inflection is a point ${𝒙}_{\mathbf{0}}$ where \boldsymbol{f''(x_0) = 0}. Once again, not all such points need be points of inflection, as is seen from $y=x^4$.

If the tests we have mentioned fail to determine whether a point is a relative maximum or minimum, or a point of inflection, one should simply test that point by referring to the definitions of these terms. There are tests involving higher derivatives, but we shall not discuss them.

Now let us consider the general problem of tracing the graph of a function $y=f(x)$ with some attention to its critical points and in a relatively economical manner. First of all, we should observe what the domain of the function is, i.e., for what values of $x$ it is defined. For instance, $f(x)=\frac{1}{x-1}$ is defined for $x\ne 1,f(x)=\sqrt{x}$ for all $x\ge 0$. We should also observe what its range is, i.e., what values $f(x)$ can take on. For example, the range of $y=\sin x$ is $-1\le y\le 1$, while that of $y=\sqrt{x}$ is all $y\ge 0$ and that of $y=x^3$ is all numbers. Next we should determine a few points of the graph, and one usually finds those where it crosses the axes. It will of course cross the y-axis when $x=0,y=f(0)$. Sometimes finding where it crosses the x-axis, i.e., where $f(x)=0$, is considerably harder, and must be estimated. The totality of such points are called the intercepts of the graph.

Then one should determine the turning points, i.e., relative maxima and minima and points of inflection. By the location of the maxima and minima and by the behavior of $f(x)$ for $x$ very far positive and very far negative, it is possible to see in what direction the curve crosses its tangent at the points of inflection. It is also useful to determine in what regions f''(x) > 0 and f''(x) < 0, to see in what direction the curve is concave. In this process, some points may occur where the first or second derivative does not exist. They should be plotted separately, and special attention should be paid to the behavior of the curve near them.

Finally, one should look for the asymptotes of the curve, which are, roughly speaking, lines approached by the curve. Vertical asymptotes will often occur at points where the function is undefined; for example, the line $x=1$ is a vertical asymptote to $y=\frac{1}{x-1}$. Horizontal asymptotes will occur when $f(x)$ approaches a constant as $x$ goes very far positive or very far negative. The line $y=0$ is, for instance, a horizontal asymptote to $y=\frac{1}{x-1}$, here both for $x$ far negative and far positive. A suspected oblique asymptote can be treated by the test for horizontal asymptotes by taking its equation $y=mx+b$ and subtracting from $f(x)$ to get the new function $g(x)=f(x)-(mx+b)$. Then $y=mx+b$ is an asymptote for $f(x)$ if and only if $y=0$ is an asymptote for $g(x)$.

By these considerations, one can make a sketch which will reveal the shape of the curve, since the behavior between the critical points can be plotted smoothly with attention only to the concavity in these regions. As an example, let us consider $f(x)=x^{2/3}(1-x)$. This is defined for all $x$, becomes far negative as $x$ becomes positive, far positive as $x$ goes far negative, and takes on all values between. Thus its range is all numbers. We see that $f(0)=0,f(1)=0$. Thus the intercepts are $(0,0)$ and $(1,0)$. Now f'(x) = \frac{2}{3}x^{-1/3}(1 - x) - x^{2/3}, where this is defined. We see from the first term that f'(0) is undefined. But for $x$ either slightly positive or slightly negative, . Therefore $x=0$ is a relative minimum. In the remaining cases, setting f'(x) = 0 gives

\begin{aligned} 0 &= \frac{2}{3}x^{-1/3}(1 - x) - x^{2/3} \\ &= \frac{2}{3}x^{-1/3} - \frac{2}{3}x^{2/3} - x^{2/3} \\ &= \frac{2}{3}x^{-1/3} - \frac{5}{3}x^{2/3} \\ &= \frac{2}{3} - \frac{5}{3}x, \end{aligned}

or $x=\frac{2}{5}$ is the only solution of f'(x) = 0. Now

\begin{aligned} f''(x) &= -\frac{2}{9}x^{-4/3}(1 - x) - \frac{2}{3}x^{-1/3} - \frac{2}{3}x^{-1/3} \\ &= -\frac{2}{9}x^{-4/3} - \frac{10}{9}x^{-1/3} \\ &= -\frac{2}{9}x^{-4/3}(1 + 5x), \quad x \neq 0 . \end{aligned}

Thus f''(\frac{2}{5}) < 0, and $x=\frac{2}{5}$ is a relative maximum. If f''(x) = 0, then $x=-\frac{1}{5}$, and f''(x) > 0 for x < -\frac{1}{5}, f''(x) < 0 for . Since the curvature changes sign at $-\frac{1}{5},x=-\frac{1}{5}$ is a point of inflection, the graph is concave upward for , and concave downward for . There are no apparent asymptotes. In the critical region, then, the curve looks like this:

An equation for the tangent to the graph of $y=f(x)$ has been given above in the discussion of concavity. However, let us derive this in parametric form, as well as the equation of the normal, or perpendicular, to $y=f(x)$ at a point. A vector $A$ parallel to the tangent line is found by taking any vector $A=(a_1,a_2)$ such that the ratio $\frac{a_2}{a_1}$ of its components is equal to the slope. At the point $x_0$, this slope is f'(x_0), and we can take a_1 = 1, a_2 = f'(x_0). The point of the curve is $(x_0,f(x_0))$, and so the parametric equation of the tangent line at $(x_0,f(x_0))$ is

(x, y) = (x_0, f(x_0)) + t(1, f'(x_0)) .

Solving for $t$ gives $t=x-x_0$; now y = f(x_0) + t f'(x_0), or

\begin{aligned} y &= f(x_0) + (x - x_0)f'(x_0) \\ &= x f'(x_0) + (f(x_0) - x_0 f'(x_0)), \end{aligned}

the equation we have used before.

A vector parallel to the normal line at $(x_0,f(x_0))$ must be perpendicular to (1, f'(x_0)). Let $(b_1,b_2)$ be such a vector. Then

b_1 + b_2 f'(x_0) = 0, \quad \text{or} \quad \frac{b_1}{b_2} = -f'(x_0) .

In particular, we can choose b_2 = 1, b_1 = -f'(x_0). The normal line is then

(x, y) = (x_0, f(x_0)) + t(-f'(x_0), 1) .

EXERCISES