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Concavity and Inflection · Tutorial 811 of 1495

Concavity Under Vertical and Horizontal Transformations

Use a single transformation formula to determine whether concavity is preserved or reversed and where an inflection point moves.

Intermediate 10 min read

What You'll Learn

  • Write translations, reflections, and nonzero scalings in a common transformation formula.
  • Derive how the second derivative changes under horizontal and vertical transformations.
  • Determine which transformations preserve or reverse concavity.
  • Map an inflection point to its transformed coordinates.
  • Avoid confusing a horizontal reflection with a reversal of concavity.

Why Transformations Matter for Concavity

A graph can be shifted, reflected, or stretched without changing its basic shape in arbitrary ways. In particular, these transformations have predictable effects on where the graph is concave up or concave down, and on the coordinates of its inflection points. Knowing those effects lets you analyze a transformed function without starting its concavity analysis from scratch.

The key distinction is between changing the graph vertically and changing it horizontally. A vertical reflection reverses concavity: it turns concave up into concave down and vice versa. A horizontal reflection does not reverse the concavity type at corresponding points, although it swaps which side of the graph corresponds to the left or right side of the original.

A General Transformation Formula

Let \(f\) be a function, and consider the transformed function

$$ g(x)=A f\bigl(B(x-H)\bigr)+K, $$

where \(A\) and \(B\) are nonzero constants and \(H\) and \(K\) are constants. The transformed function is defined at each \(x\) for which \(B(x-H)\) belongs to the domain of \(f\).

In this formula, \(H\) gives a horizontal translation, and \(K\) gives a vertical translation. The factor \(B\) controls horizontal scaling and reflection: its magnitude changes horizontal scale, while a negative value also reflects the graph across the vertical axis relative to its horizontal center. The factor \(A\) controls vertical scaling and reflection: a negative value reflects the graph across the horizontal axis.

Definition: An inflection point is a point on a function’s graph where the function changes concavity. In using this definition, concavity must change across the point; merely having \(f''(c)=0\), or having \(f''(c)\) fail to exist, does not by itself establish an inflection point.

For a twice-differentiable function, the sign of \(f''\) identifies concavity on an interval: \(f''(x)>0\) means concave up there, and \(f''(x)<0\) means concave down there. A change in the sign of \(f''\) across a point is a useful way to confirm a change in concavity when the second derivative exists on both sides.

How the Second Derivative Changes

Differentiate \(g(x)=A f(B(x-H))+K\) using the chain rule. The constants \(A\) and \(K\) do not depend on \(x\), and the inner function \(B(x-H)\) has derivative \(B\). Differentiating a second time introduces another factor of \(B\).

$$ g'(x)=AB f'\bigl(B(x-H)\bigr) $$
$$ g''(x)=AB^2 f''\bigl(B(x-H)\bigr). $$

This formula requires \(f\) to be twice differentiable at the input \(B(x-H)\). The factor \(B^2\) is positive because \(B\ne 0\). Therefore, horizontal scaling—including horizontal reflection—does not itself reverse the sign of the second derivative at corresponding points. The sign is multiplied by \(A\): positive \(A\) preserves the concavity type, while negative \(A\) reverses it.

Theorem: For \(g(x)=A f(B(x-H))+K\) with \(A\ne 0\) and \(B\ne 0\), the second derivative, wherever it exists, is \(g''(x)=AB^2 f''(B(x-H))\). Thus \(A>0\) preserves concavity at corresponding points, and \(A<0\) reverses it. The horizontal factor \(B\) changes which \(x\)-values correspond, but not the concavity type when \(A>0\).

Deriving the Inflection-Point Coordinates

Suppose \((u,f(u))\) is an inflection point of \(f\). To find the corresponding horizontal coordinate on the transformed graph, set the input to \(f\) equal to \(u\): \(B(x-H)=u\). Since \(B\ne 0\), this equation has the unique solution \(x=H+u/B\). The transformed vertical coordinate is obtained by applying the vertical transformation to \(f(u)\).

$$ (x,y)=\left(H+\frac{u}{B},\; A f(u)+K\right). $$

This point is an inflection point of \(g\). To see why, note that the horizontal change of variables \(u=B(x-H)\) maps neighborhoods of the original point to neighborhoods of the transformed point. If \(B<0\), it reverses their left-right order; it does not remove the change in concavity. Multiplication by \(A>0\) preserves concavity, while multiplication by \(A<0\) exchanges concave up and concave down. Either way, a change in concavity remains a change in concavity. The transformed graph is therefore an inflection-point image, not merely a point where its second derivative might be zero.

The argument also explains why a horizontal reflection can swap which side is concave up without changing the concavity type at corresponding locations. The two sides are exchanged, while the concavity on each matching piece remains the same if \(A>0\).

Worked Examples

Worked Example: A Shift and a Positive Vertical Stretch

Let \(f(x)=x^3\), and define \(g(x)=2f(x-1)+3\). Find the concavity of \(g\) and its inflection point.

Here \(A=2\), \(B=1\), \(H=1\), and \(K=3\). The original derivatives are \(f'(x)=3x^2\) and \(f''(x)=6x\). The transformation formula gives

$$ g''(x)=2(1)^2 f''(x-1) =2\bigl(6(x-1)\bigr) =12(x-1). $$

If \(x<1\), then \(g''(x)<0\), so \(g\) is concave down. If \(x>1\), then \(g''(x)>0\), so \(g\) is concave up. The original inflection point is \((0,0)\). Its transformed coordinates are \(x=1+0/1=1\) and \(y=2(0)+3=3\).

Check by substitution: \(g(1)=2f(0)+3=2(0)+3=3\), so \((1,3)\) is on the transformed graph. Also, \(g''(1)=12(1-1)=0\), and the sign changes from negative to positive across \(x=1\). Thus the transformed point is indeed an inflection point.

Worked Example: A Horizontal Reflection and Stretch

Let \(f(x)=x^3\), and set \(g(x)=f(-2x)\). Determine the concavity intervals and the inflection point.

In the general formula, \(A=1\), \(B=-2\), \(H=0\), and \(K=0\). Since \(f''(u)=6u\),

$$ g''(x)=(1)(-2)^2 f''(-2x) =4\bigl(6(-2x)\bigr) =-48x. $$

For \(x<0\), \(g''(x)>0\), so \(g\) is concave up. For \(x>0\), \(g''(x)<0\), so \(g\) is concave down. The original inflection point \((0,0)\) maps to \((0,0)\): its new horizontal coordinate is \(0+0/(-2)=0\), and its vertical coordinate is \(f(0)=0\).

Check by writing the function explicitly: \(g(x)=(-2x)^3=-8x^3\). Then \(g'(x)=-24x^2\) and \(g''(x)=-48x\), matching the formula. Also \(g(0)=0\), and the second derivative changes sign at \(0\). The horizontal reflection has exchanged which side of the original point supplies each side of the transformed graph, but it has not added a vertical reflection.

Worked Example: A General Transformation of a Nonzero Inflection Point

Let \(f(x)=(x-1)^3+2\), and define \(g(x)=-2f(3(x-4))+1\). Find the transformed inflection point and concavity on either side.

First, \(f'(x)=3(x-1)^2\) and \(f''(x)=6(x-1)\). The second derivative changes from negative to positive at \(x=1\), so the original inflection point is \((1,2)\). The parameters are \(A=-2\), \(B=3\), \(H=4\), and \(K=1\). The coordinate formula gives

$$ x=4+\frac{1}{3}=\frac{13}{3}, \qquad y=(-2)(2)+1=-3. $$

To find the concavity, write the second derivative in terms of \(x\):

$$ g''(x)=(-2)(3)^2 f''\bigl(3(x-4)\bigr) =-18\left(6\bigl(3(x-4)-1\bigr)\right) =-108(3x-13). $$

If \(x<13/3\), then \(3x-13<0\), so \(g''(x)>0\) and \(g\) is concave up. If \(x>13/3\), then \(3x-13>0\), so \(g''(x)<0\) and \(g\) is concave down. The original graph was concave down to the left of \(1\) and up to its right; the horizontal transformation preserves the corresponding concavity type, while the negative vertical factor reverses it.

Check the point directly. At \(x=13/3\), the input to \(f\) is \(3(13/3-4)=1\). Hence \(g(13/3)=-2f(1)+1=-2(2)+1=-3\), as predicted. Also \(g''(13/3)=-108(13-13)=0\), and its sign changes from positive to negative there.

A Common Mistake: Giving a Horizontal Reflection a Negative Second-Derivative Factor

A common error is to assume that a negative horizontal scale reverses concavity. For example, one might incorrectly differentiate \(g(x)=f(-2x)\) twice and claim that \(g''(x)=-4f''(-2x)\). That is not the chain rule: the inner derivative \(-2\) appears twice, so the product is \((-2)(-2)=4\), not \(-4\).

$$ g'(x)=-2f'(-2x), \qquad g''(x)=(-2)(-2)f''(-2x)=4f''(-2x). $$

For \(f(x)=x^3\), this gives \(g''(x)=4\cdot 6(-2x)=-48x\), not \(48x\). A horizontal reflection changes which original side corresponds to each new side. It does not, by itself, flip concave up to concave down at the corresponding locations. A negative vertical factor is what reverses the concavity type.

Using the Transformation Rules Reliably

For a transformed function \(g(x)=A f(B(x-H))+K\), identify \(A\), \(B\), \(H\), and \(K\) before drawing conclusions. Then use \(AB^2 f''(B(x-H))\) for the second derivative and solve \(B(x-H)=u\) to map an original inflection coordinate \(u\). This keeps the concavity calculation and the coordinate calculation separate, reducing sign and placement errors.

The nonzero conditions matter. If \(B=0\), the input to \(f\) is constant, so the graph is constant wherever defined and the horizontal coordinate map cannot be inverted. If \(A=0\), the transformed graph is also constant, regardless of the shape of \(f\). Neither case is an invertible scaling of the graph, so the transformation and inflection correspondence above do not apply.

These rules are useful when comparing families of graphs or interpreting a function after a change of scale or origin. They also provide a quick check on a sketch: a vertical reflection must exchange concavity types, while a horizontal reflection must exchange left and right correspondence. Translations change locations, but do not alone alter concavity.

Check Your Understanding

Use the transformation formula and explain each sign or coordinate choice.

  1. For \(g(x)=3f(x+2)-1\), does concavity at corresponding points stay the same or reverse?
  2. If \(f\) has an inflection point at \((u,f(u))\), what are its coordinates on \(g(x)=f(-4(x-1))+5\)?
  3. For \(g(x)=-f(2x)\), which part of the transformation reverses concavity: the negative horizontal factor, the negative vertical factor, or both?
  4. If \(g''(x)=AB^2f''(B(x-H))\), why is the sign of \(B\) not itself a concavity-reversal factor?
  5. Why is \(f''(c)=0\) alone insufficient to conclude that \(c\) is an inflection point?

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