Curve Intersections and Which Curve Is Higher, foundations for calculus lesson

Two taxi services price a ride in different ways, one with a high base fare and a low rate per kilometer, the other the reverse. For short rides one is cheaper, for long rides the other, and at one distance they cost the same. That break-even distance is where two graphs cross, and the question of which plan is cheaper is the question of which graph is lower. Calculus asks the same two questions whenever it finds the area between two curves.

Key terms

Curve intersection and common point

Definition A point $(a, b)$ is a common point, or intersection point, of two curves when it lies on both of them. For the graphs of functions $y = f(x)$ and $y = g(x)$ this means that $a$ is in the domain of $f$ and in the domain of $g$ , and

f(a) = g(a) = b.

The intersection of the two curves is the set of all such points.

In plain words The two graphs pass through the very same spot.

Example The graphs of $f(x) = 2x$ and $g(x) = x + 3$ meet where $2x = x + 3$ , that is at $x = 3$ . Both functions give $6$ there, so the common point is $(3, 6)$ .

Same input and same output

Definition At an intersection of two function graphs, the input is the same for both functions and the output is the same for both functions. Equal outputs at different inputs do not give a common point, and a common input with different outputs does not either.

In plain words Both coordinates must agree. Hitting the same height at different places is not a meeting.

Example For $f(x) = 2x$ and $g(x) = x + 3$ , we have $f(2) = 4$ and $g(1) = 4$ . The output $4$ is shared, but the points $(2, 4)$ and $(1, 4)$ are different points, so neither is an intersection. At the common input $x = 1$ the outputs are $2$ and $4$ , again no meeting. The only common point is $(3, 6)$ .

Simultaneous equations

Definition The intersection points of $y = f(x)$ and $y = g(x)$ are the solutions $(x, y)$ of the two equations taken together. Eliminating $y$ gives the single equation

\begin{gathered} f(x) = g(x), \\[0.4em] x \in \operatorname{dom}(f) \cap \operatorname{dom}(g). \end{gathered}

Each solution $x = a$ gives the intersection point $(a, f(a))$ . Solutions outside the common domain do not count.

In plain words Set the two outputs equal, solve for $x$ , then use either formula to get the height.

Example Take $f(x) = x^2$ and $g(x) = x + 2$ , both defined for all real $x$ . Setting them equal,

\begin{aligned} & x^2 = x + 2 \\[0.4em] \iff {} & x^2 - x - 2 = 0 \\[0.4em] \iff {} & (x - 2)(x + 1) = 0. \end{aligned}

So $x = 2$ or $x = -1$ . The intersection points are $(2, 4)$ and $(-1, 1)$ , and each one checks in both formulas.

Extraneous intersections

Definition Solving $f(x) = g(x)$ sometimes uses a step that is not reversible, such as squaring both sides. Such a step can produce candidates that are not solutions of the original equation. A candidate is a true intersection only if it lies in both domains and makes $f(x) = g(x)$ hold exactly. A candidate that fails is called extraneous and is discarded.

In plain words Squaring makes $2 = -2$ look like $4 = 4$ . After squaring, plug every candidate back into the original equation.

Example Find where $y = \sqrt{x}$ meets the line $y = x - 2$ . Squaring $\sqrt{x} = x - 2$ gives

\begin{aligned} & x = x^2 - 4x + 4 \\[0.4em] \iff {} & x^2 - 5x + 4 = 0 \\[0.4em] \iff {} & (x - 1)(x - 4) = 0. \end{aligned}

Check each candidate in the original equation. At $x = 1$ the root gives $1$ while the line gives $-1$ , so $x = 1$ is extraneous. At $x = 4$ both give $2$ . The only intersection is $(4, 2)$ . The false candidate appeared because $\sqrt{x}$ is never negative, while $x - 2$ is negative at $x = 1$ .

Vertical difference

Definition For two functions $f$ and $g$ , the vertical difference is the function

\begin{gathered} d(x) = f(x) - g(x), \\[0.4em] x \in \operatorname{dom}(f) \cap \operatorname{dom}(g). \end{gathered}

Its value is the signed vertical distance from the graph of $g$ up to the graph of $f$ at the input $x$ . The intersections of the two graphs occur exactly at the zeros of $d$ .

In plain words Stand at one input and measure from the graph of $g$ to the graph of $f$ . The size of the number is the gap, and its sign tells you which graph is on top.

Example For $f(x) = x + 2$ and $g(x) = x^2$ ,

\begin{aligned} d(x) &= x + 2 - x^2 \\[0.4em] &= -(x - 2)(x + 1). \end{aligned}

At $x = 0$ the gap is $d(0) = 2$ , so the line is $2$ units above the parabola there. The zeros $x = -1$ and $x = 2$ are the intersections found before.

Which curve is higher

Definition At an input $x$ in both domains, the graph of $f$ is higher than the graph of $g$ when $f(x) - g(x) > 0$ , lower when $f(x) - g(x) < 0$ , and the graphs meet when the difference is $0$ . Higher and lower are statements about one input at a time.

In plain words Subtract and look at the sign. Positive means the first function is on top.

Example For $f(x) = x + 2$ and $g(x) = x^2$ the difference at $x = -2$ is $d(-2) = -4$ , so the parabola is higher there. At $x = 0$ it is $2$ , so the line is higher. Later, the area between two curves is built from the top function minus the bottom function, so getting this sign right matters.

Interval splitting

Definition Suppose $d = f - g$ is continuous on an interval and has no zeros there. Then $d$ keeps one sign on the whole interval, by the intermediate value theorem, since a continuous function that took a positive and a negative value on an interval would have to equal $0$ somewhere between them. So when $f$ and $g$ are continuous on their domains, find all zeros of $d$ and all points where the common domain breaks into separate intervals, split the common domain at those points, and test one input in each piece. A sampled picture or a table of values alone does not prove where the graphs cross or which is higher, because two crossings can hide between sample points.

In plain words The graphs can only swap places where they meet or where something breaks. Between those points, one test value tells you the whole story.

Example For $d(x) = x + 2 - x^2$ the zeros are $-1$ and $2$ .

Interval	Test input	Sign of $d$	Higher graph
$(-\infty, -1)$	$-2$	negative	parabola
$(-1, 2)$	$0$	positive	line
$(2, \infty)$	$3$	negative	parabola

Sampling alone can mislead. For $f(x) = x^2$ and $g(x) = 3x - 2.16$ the differences at $x = 0, 1, 2, 3$ are $2.16$ , $0.16$ , $0.16$ and $2.16$ , all positive. Yet the difference factors as $(x - 1.2)(x - 1.8)$ , so the graphs cross twice and the line is higher on $(1.2, 1.8)$ .

Check yourself

Find the intersection points of $f(x) = x^2 - 2x$ and $g(x) = x + 4$ , and decide which graph is higher between them.

Show the answer

Setting the outputs equal gives $x^2 - 3x - 4 = 0$ , so $(x - 4)(x + 1) = 0$ and $x = 4$ or $x = -1$ . The points are $(4, 8)$ and $(-1, 3)$ . At $x = 0$ the difference $g(x) - f(x)$ equals $4$ , which is positive, so the line is higher on $(-1, 4)$ .

Find every intersection of $y = \sqrt{x + 2}$ and $y = x$ .

Show the answer

Squaring gives $x + 2 = x^2$ , so $(x - 2)(x + 1) = 0$ and the candidates are $2$ and $-1$ . At $x = -1$ the root gives $1$ and the line gives $-1$ , so that candidate is extraneous. At $x = 2$ both give $2$ . The only intersection is $(2, 2)$ .

Where is the graph of $f(x) = x^3$ higher than the graph of $g(x) = x$ ?

Show the answer

The difference factors as $x^3 - x = x(x - 1)(x + 1)$ , with zeros $-1$ , $0$ and $1$ . Testing $-2$ , $-0.5$ , $0.5$ and $2$ gives the signs negative, positive, negative and positive. So the cubic is higher on $(-1, 0) \cup (1, \infty)$ .

Curve Intersections and Which Curve Is Higher

Key terms

Curve intersection and common point

Same input and same output

Simultaneous equations

Extraneous intersections

Vertical difference

Which curve is higher

Interval splitting

Check yourself

How this lesson connects

Builds on

Where this leads in Calculus