When One Point Has Two Approaching Slopes
A smooth-looking graph often suggests a single direction as you move closer to a point. At a corner, the direction from the left can differ from the direction from the right. Secant slopes make this difference precise: the slopes of lines joining a fixed point on the graph to nearby points may approach one value from one side and another value from the other side.
When those two values differ, there is no single slope that describes the graph's tangent direction at the point. The graph of \(y=|x|\) gives the simplest example. Its two straight sides meet at the origin, but the left and right sides have different slopes.
One-Sided Tangent Slopes
Let \(f\) be a function defined on an interval around \(c\). For a nearby point \(x\ne c\), the secant slope between \((c,f(c))\) and \((x,f(x))\) is
As \(x\) approaches \(c\) from the left or right, these secant slopes may approach a finite value. If they do, those values describe the left-hand and right-hand tangent slopes.
The notation \(x\to c^-\) means that \(x\) approaches \(c\) using values less than \(c\); \(x\to c^+\) means it approaches using values greater than \(c\). These are separate approaches. A single tangent slope requires the secant slopes to settle on the same finite value from both sides.
The Basic Corner: \(y=|x|\)
For \(f(x)=|x|\), the point at the corner is \((0,0)\), since \(f(0)=0\). To find the left-hand secant slope, take \(x<0\). Then \(|x|=-x\), so
This quotient is exactly \(-1\) for every \(x<0\), so its limit as \(x\to0^-\) is \(-1\). On the right, \(x>0\), so \(|x|=x\), and
This quotient is exactly \(1\) for every \(x>0\), so its limit as \(x\to0^+\) is \(1\). Because \(-1\ne1\), the two sides do not approach a common slope.
Each side is a straight line: to the left of the origin the graph lies on \(y=-x\), and to the right it lies on \(y=x\). Those lines meet at the corner, but neither one gives a single direction that matches both sides. Calling either line “the” tangent at the corner would ignore the other side.
Why Unequal Limits Rule Out One Slope
The conclusion follows from what it means for a two-sided limit to exist. If the secant slopes approached one finite value \(m\) as \(x\to c\), then they would have to approach \(m\) whether \(x\) came from below or above. In particular, each one-sided limit would have to equal \(m\).
Therefore, if the one-sided limits exist but differ, no such \(m\) can describe both approaches. This is not a failure to calculate a slope accurately: it is evidence that one unique tangent slope does not exist. For \(y=|x|\), a proposed common value would have to equal \(-1\) from the left and \(1\) from the right, which is impossible.
Use \(\frac{f(x)-f(c)}{x-c}\) for \(x\ne c\).
Find the limit using \(x<c\), if it exists.
Find the limit using \(x>c\), if it exists.
Different finite values mean there is no unique tangent slope at \(c\).
Worked Examples
Worked Example: A Shifted Absolute-Value Corner
Consider \(f(x)=|x-2|+1\). Find the one-sided tangent slopes at \(x=2\) and decide whether there is a unique tangent slope.
First, \(f(2)=|2-2|+1=1\), so the point is \((2,1)\). If \(x<2\), then \(x-2<0\) and \(|x-2|=2-x\). The secant slope is
Thus the left-hand tangent slope is \(-1\). If \(x>2\), then \(|x-2|=x-2\), and
The right-hand tangent slope is \(1\). They differ, so there is no unique tangent slope at \(x=2\). As a check, the two sides of the graph are \(y=3-x\) for \(x<2\) and \(y=x-1\) for \(x>2\); both give \(y=1\) at \(x=2\), and their slopes are \(-1\) and \(1\), respectively.
Worked Example: A Corner with Unequal Side Steepness
Let \(g(x)=3-|2x+4|\). Find the one-sided tangent slopes at \(x=-2\).
At \(x=-2\), \(2x+4=0\), so \(g(-2)=3\). When \(x<-2\), \(2x+4<0\), hence \(|2x+4|=-(2x+4)\). Therefore
The left-hand tangent slope is \(2\). When \(x>-2\), \(2x+4>0\), so \(|2x+4|=2x+4\). Then
The right-hand tangent slope is \(-2\), so no unique tangent slope exists at \(x=-2\). As a check, the graph is \(2x+7\) on the left and \(-2x-1\) on the right. At \(x=-2\), both expressions equal \(3\), and their slopes are \(2\) and \(-2\), as found from the secants.
Worked Example: A Piecewise Linear Corner
Define \(p(x)=x+1\) for \(x<1\), and \(p(x)=4-2x\) for \(x\ge1\). Find the one-sided tangent slopes at \(x=1\).
The value at the point uses the second rule: \(p(1)=4-2(1)=2\). For \(x<1\), use the first rule to compute the secant slope:
The left-hand tangent slope is \(1\). For \(x>1\), use the second rule:
The right-hand tangent slope is \(-2\). Because \(1\ne-2\), there is no unique tangent slope at \(x=1\). Checking the graph's two formulas, the left branch \(y=x+1\) and right branch \(y=4-2x\) both give \(y=2\) at \(x=1\), with slopes \(1\) and \(-2\). The branches meet, but their directions differ.
A Common Mistake: Ignoring the Side
For \(f(x)=|x|\), a tempting but incorrect calculation is
This equality is true only when \(x>0\). If \(x<0\), then \(|x|=-x\), giving
So the quotient is not \(1\) on both sides. Its left-hand limit is \(-1\), and its right-hand limit is \(1\). The correction is to use the appropriate expression for the absolute value on each side before taking a limit. This is essential at a corner, where the two sides can follow different rules or have different directions.
Another useful distinction is that a corner need not be a break in the graph. In the examples above, the two branches meet at the same point. The failure is specifically the mismatch between their approaching slopes. The test concerns the one-sided secant slopes, not merely whether the graph has a point at \(c\).
Why Corners Matter
A corner marks a point where the graph changes direction abruptly. Knowing that the two one-sided slopes differ lets us describe this local behavior precisely, rather than guessing from the appearance of a sketch. The same check works for absolute-value expressions and piecewise-defined graphs: identify the point, form the secant quotient on each side, and compare the limits.
The conclusion is limited to the situation considered here: two finite one-sided slopes that approach different values. If the secant slopes behave in some other way, that requires a separate analysis. In particular, the important point in this tutorial is the unequal finite values at a corner, not every possible way a tangent slope might fail to exist.
Summary
- A one-sided tangent slope is the limit of secant slopes as the nearby point approaches from one specified side.
- At a corner such as \(y=|x|\) at the origin, the left and right slopes are \(-1\) and \(1\).
- If the left- and right-hand tangent slopes exist but differ, there is no unique tangent slope.
- For absolute values and piecewise formulas, use the correct expression on each side before simplifying the secant quotient.
Check Your Understanding
For each question, compare the two sides when the graph has a corner.
- For \(f(x)=|x|\), what are the left- and right-hand tangent slopes at \(x=0\)?
- For \(q(x)=|x+3|\), find the corner's \(x\)-coordinate and the two one-sided tangent slopes there.
- Suppose a function has left-hand tangent slope \(4\) and right-hand tangent slope \(4\) at a point. Do these values establish a mismatch of the kind seen at a corner?
- For \(r(x)=|x-1|+2\), compute the secant quotient on each side of \(x=1\) and decide whether a unique tangent slope exists.
- Why is the calculation \(|x|/x=1\) invalid when \(x<0\)?
