When Secant Slopes Grow Without Bound
A tangent does not always have a finite slope. A curve can become increasingly steep near a point, so that the slopes of secant lines grow without bound as their second point approaches the point of interest. Geometrically, the secant lines can approach a vertical line.
The curve \(y=\sqrt[3]{x}\) provides a clear example. It passes through the origin, and its shape near the origin is not a corner: from both sides, nearby secants turn toward the same vertical direction. We will calculate those slopes and justify what “grow without bound” means.
Secant Slopes and Vertical Tangents
For a graph \(y=f(x)\), choose two points on the curve, \((a,f(a))\) and \((x,f(x))\), with \(x\ne a\). The slope of the secant line through them is the change in output divided by the change in input.
As \(x\) approaches \(a\), the second point approaches the first. Earlier material on tangent slopes describes how secant slopes can approach a finite number, producing an ordinary tangent slope. Another possibility is that the secant slopes become unbounded.
An unbounded slope is not a real-number slope. A vertical line itself has no slope in the usual form \(\frac{\text{change in }y}{\text{change in }x}\), because its horizontal change is zero. Thus, a vertical tangent is different from an ordinary tangent with a very large but finite slope.
Worked Example: Secants on the Cube-Root Curve
Worked Example: Secants on the Cube-Root Curve
Let \(f(x)=\sqrt[3]{x}\), and find the secant slope between the origin and the point corresponding to \(x=8\).
The first point is \((0,f(0))=(0,0)\). The second point is \((8,f(8))=(8,2)\), because \(\sqrt[3]{8}=2\). Therefore,
The secant line through \((0,0)\) with slope \(\frac14\) is \(y=\frac14x\). Check: at \(x=8\), this line gives \(y=\frac14(8)=2\), so it passes through both selected points.
This one secant is not enough to determine the tangent direction at the origin. We need to see what happens as the second point moves closer, including from both the left and the right.
Deriving the Secant Slope Near the Origin
For \(x\ne0\), the secant slope from \((0,0)\) to \((x,\sqrt[3]{x})\) is
Since \(x=(\sqrt[3]{x})^3\), and \(x\ne0\), we can simplify:
The final expression is positive on both sides of zero. This matters: for \(x<0\), both the vertical change \(\sqrt[3]{x}\) and the horizontal change \(x\) are negative, so their ratio is positive.
Worked Example: Comparing Left- and Right-Hand Secants
Calculate the secant slopes at \(x=\frac18\) and \(x=-\frac18\), then compare them with slopes closer to the origin.
For \(x=\frac18\), the point on the curve is \((\frac18,\frac12)\), since \(\sqrt[3]{\frac18}=\frac12\). Its secant slope from the origin is
For \(x=-\frac18\), the point is \((-\frac18,-\frac12)\). The slope is
Now take \(x=\frac{1}{1000}\). Since \(\sqrt[3]{\frac{1}{1000}}=\frac{1}{10}\), the slope is \(\frac{1/10}{1/1000}=100\). At \(x=-\frac{1}{1000}\), the point is \((-\frac{1}{1000},-\frac{1}{10})\), and the slope is \(\frac{-1/10}{-1/1000}=100\).
Check each calculation using \(m(x)=1/|x|^{2/3}\): at \(|x|=\frac18\), this gives \(1/(1/4)=4\); at \(|x|=\frac{1}{1000}\), it gives \(1/(1/100)=100\). The direct quotients and the simplified formula agree on both sides.
Why the Slopes Become Unbounded
Examples suggest that the slopes grow as \(x\) gets closer to zero. The exact formula proves this, and the following argument gives a precise meaning to “grow without bound.”
For \(x\ne0\), \(m(x)=1/|x|^{2/3}\), so \(m(x)>0\) on both sides of zero.
Let \(M>0\) be any number, however large. We want the slope to exceed \(M\).
Set \(\delta=M^{-3/2}\). If \(0<|x|<\delta\), then \(|x|^{2/3}<\delta^{2/3}=M^{-1}\).
Taking reciprocals of positive quantities gives \(m(x)=1/|x|^{2/3}>M\). This holds for positive and negative \(x\).
Because this argument works for every \(M>0\), no finite upper bound can contain the nearby secant slopes. In the standard limit language, \(m(x)\) tends to \(+\infty\) from both sides as \(x\) tends to zero.
Why This Means the Tangent Direction Is Vertical
The secant line from the origin to \((x,\sqrt[3]{x})\) has equation \(y=m(x)x\). A direction vector for a line of slope \(m(x)\) is \((1,m(x))\). As \(m(x)\) becomes positive and unbounded, these direction vectors point increasingly close to the vertical direction. For example, after dividing by \(m(x)>0\), the same direction is represented by \((1/m(x),1)\); its horizontal component tends to zero while its vertical component stays equal to one.
The secant lines also pass through the origin, so their limiting vertical line is \(x=0\). This is the vertical tangent to \(y=\sqrt[3]{x}\) at the origin. Notice the distinction: the secant slopes do not approach a finite real number, but the secant directions do approach a definite vertical direction.
Worked Example: A Shifted Cube-Root Curve
Consider \(g(x)=2+\sqrt[3]{x-3}\). Find the secant slopes from \((3,2)\) at \(x=3+\frac18\) and \(x=3-\frac18\), and identify the limiting vertical line.
First, \(g(3)=2+\sqrt[3]{0}=2\), so \((3,2)\) is on the curve. For \(x=3+\frac18=\frac{25}{8}\), we have \(x-3=\frac18\), so \(g(x)=2+\frac12=\frac52\). The secant slope is
For \(x=3-\frac18=\frac{23}{8}\), we have \(x-3=-\frac18\), so \(g(x)=2-\frac12=\frac32\). The secant slope is
More generally, writing \(h=x-3\), the slope from \((3,2)\) is \(\frac{\sqrt[3]{h}}{h}=\frac{1}{|h|^{2/3}}\) for \(h\ne0\). Thus the slopes grow without bound from either side as \(h\) approaches zero. The secant lines approach the vertical line \(x=3\).
Check: both computed points lie on \(g\), and substituting their coordinates into the secant formula gives slope \(4\) in each case. The shifted curve has the same local slope behavior as the original cube-root curve, with the vertical tangent shifted to \(x=3\).
A Common Sign Mistake
A frequent error is to assume that a point to the left of the origin must give a negative secant slope. That conclusion confuses the signs of the separate changes with the sign of their ratio. For the cube-root curve at \(x=-\frac18\), the incorrect calculation might be written as
The correction is that a negative divided by a negative is positive:
This sign is consistent with the exact formula \(m(x)=1/|x|^{2/3}>0\). On both sides, the secant line rises as \(x\) increases, and its direction becomes more nearly vertical near the origin.
What to Look For
To investigate a possible vertical tangent using secants, fix a point on the curve and compute the slopes to nearby points. Check the two sides separately: the slopes may have different signs, and a vertical tangent does not require them to behave identically. Then ask whether the magnitudes become arbitrarily large and whether the secant directions approach a vertical line.
For \(y=\sqrt[3]{x}\), the slopes from the origin are positive and unbounded from both sides. There is therefore a vertical tangent \(x=0\), not a finite tangent slope. The key evidence is not merely that a few slopes are large, but that the exact slope formula and the limiting argument show they exceed any prescribed positive number sufficiently close to the origin.
Check Your Understanding
Answer the following questions using the secant-slope formula and the behavior near the point.
- For \(y=\sqrt[3]{x}\), what is the secant slope from the origin to the point with \(x=\frac{1}{27}\)?
- What is the secant slope from the origin to the point with \(x=-\frac{1}{27}\), and why is its sign positive?
- In the argument for unbounded slopes, why does choosing \(\delta=M^{-3/2}\) ensure that the slope exceeds \(M\) whenever \(0<|x|<\delta\)?
- What is the equation of the vertical tangent to \(y=\sqrt[3]{x}\) at the origin?
- For \(g(x)=2+\sqrt[3]{x-3}\), what vertical line do nearby secants approach at \((3,2)\)?
