Why the Concavity of an Inverse Needs Care
Taking the inverse of a function reflects its graph across the line \(y=x\). That reflection can change how the graph bends. In particular, an increasing function and its inverse have opposite concavity when their second derivatives are nonzero, while a decreasing function and its inverse have the same concavity.
The distinction depends on the direction in which the original function moves. A formula for the second derivative of the inverse makes that dependence precise. It also explains why the sign of the original function’s second derivative, by itself, is not enough to determine the inverse’s concavity.
Conditions and the Second-Derivative Formula
Let \(f:I\to J\) be a twice differentiable, one-to-one function on an open interval \(I\), with range \(J=f(I)\). Write its inverse as \(g=f^{-1}:J\to I\). To use the formula below throughout \(I\), assume \(f'(t)\ne 0\) for every \(t\in I\). In this setting, \(f\) is strictly monotone and its inverse is twice differentiable on \(J\).
The formula evaluates the derivatives of \(f\) at \(g(x)\), the input that \(f\) maps to \(x\). The denominator is the cube of \(f'\), not its square. Its sign therefore depends on whether \(f\) is increasing or decreasing.
Deriving the Formula
The defining relation between a function and its inverse is \(f(g(x))=x\). The inverse derivative rule follows by differentiating this relation: the chain rule gives \(f'(g(x))g'(x)=1\), so, since \(f'(g(x))\ne0\), $$ g'(x)=\frac{1}{f'(g(x))}. $$
Differentiate this expression once more. Applying the chain rule to the denominator gives $$ g''(x)=-\frac{f''(g(x))g'(x)}{(f'(g(x)))^2}. $$ Now substitute \(g'(x)=1/f'(g(x))\): $$ g''(x)=-\frac{f''(g(x))}{(f'(g(x)))^2}\cdot\frac{1}{f'(g(x))} =-\frac{f''(g(x))}{(f'(g(x)))^3}. $$ This proves the formula. Each differentiation is valid because \(f\) is twice differentiable and its derivative does not vanish on \(I\).
The same derivation can be read directly from the identity \(f(g(x))=x\). Differentiating twice gives $$ f''(g(x))(g'(x))^2+f'(g(x))g''(x)=0. $$ Using \(g'(x)=1/f'(g(x))\) and dividing by \(f'(g(x))\ne0\) yields the same expression for \(g''(x)\).
How the Slope Determines Concavity
A twice differentiable function is concave up on an interval when its second derivative is positive throughout that interval, and concave down when its second derivative is negative throughout. The inverse formula compares signs at corresponding points. Since \(f'\) is continuous and never zero on an interval, its sign is constant there.
For an increasing function, \((f'(t))^3>0\), so the minus sign in the formula reverses the sign of \(f''(t)\). For a decreasing function, \((f'(t))^3<0\); dividing by this negative quantity and then applying the minus sign leaves the sign of \(f''(t)\) unchanged.
| Original function on \(I\) | Sign of \(f'\) | Concavity relationship with its inverse |
|---|---|---|
| Increasing and concave up | Positive | Inverse is concave down |
| Increasing and concave down | Positive | Inverse is concave up |
| Decreasing and concave up | Negative | Inverse is concave up |
| Decreasing and concave down | Negative | Inverse is concave down |
These statements concern intervals where the indicated derivative signs hold. If \(f''\) changes sign, the corresponding inverse curvature must be assessed on the corresponding intervals using the same formula. The inverse maps each input \(t\) to \(f(t)\), so comparisons are made between matching points, not necessarily between equal numerical inputs.
Worked Examples
Worked Example: An Increasing Function with a Concave-Down Inverse
Let \(f(t)=t^3\) for \(t>0\). This function is one-to-one, has range \((0,\infty)\), and has inverse \(g(x)=\sqrt[3]{x}\) for \(x>0\). We have $$ f'(t)=3t^2>0,\qquad f''(t)=6t>0. $$ Thus \(f\) is increasing and concave up on its domain. Apply the inverse formula at \(t=g(x)=x^{1/3}\): $$ g''(x)=-\frac{6t}{(3t^2)^3} =-\frac{6t}{27t^6} =-\frac{2}{9t^5} =-\frac{2}{9x^{5/3}}. $$ This is negative for \(x>0\), so the inverse is concave down. To check by direct differentiation, $$ g'(x)=\frac{1}{3}x^{-2/3},\qquad g''(x)=\frac{1}{3}\left(-\frac{2}{3}\right)x^{-5/3} =-\frac{2}{9x^{5/3}}. $$ At the corresponding points \(t=2\) and \(x=f(2)=8\), the formula gives \(-12/(12^3)=-1/144\), since \(f''(2)=12\) and \(f'(2)=12\). The direct expression gives \(-2/(9\cdot 8^{5/3})=-2/(9\cdot32)=-1/144\). The results agree.
Worked Example: A Decreasing Function with a Concave-Up Inverse
Let \(f(t)=e^{-t}\) for \(t\in\mathbb R\). Its derivative \(f'(t)=-e^{-t}\) is negative everywhere, so \(f\) is one-to-one and decreasing. Its range is \((0,\infty)\), and solving \(x=e^{-t}\) for \(t\) gives \(g(x)=-\ln x\), for \(x>0\). Also, $$ f''(t)=e^{-t}>0. $$ At \(t=g(x)=-\ln x\), we have \(e^{-t}=x\), so \(f'(t)=-x\) and \(f''(t)=x\). The formula gives $$ g''(x)=-\frac{x}{(-x)^3} =-\frac{x}{-x^3} =\frac{1}{x^2}>0. $$ Thus the inverse is concave up, as expected for a decreasing function with positive second derivative. Directly differentiating the inverse confirms the result: $$ g'(x)=-\frac{1}{x},\qquad g''(x)=\frac{1}{x^2}. $$ For a numerical check, take \(t=0\), so \(x=f(0)=1\). The formula gives \(-1/(-1)^3=1\), and the direct derivative gives \(g''(1)=1/1^2=1\).
Worked Example: A Decreasing Function with a Concave-Down Inverse
Let \(f(t)=-t-t^2\) for \(t>0\). Its derivative \(f'(t)=-1-2t<0\), so it is strictly decreasing. Its second derivative is \(f''(t)=-2<0\), so it is concave down. Its range is \((-\infty,0)\). Solving \(x=-t-t^2\) for the positive value of \(t\) gives $$ g(x)=\frac{-1+\sqrt{1-4x}}{2},\qquad x<0. $$ The inverse formula can be evaluated without first differentiating this explicit expression. Since \(t=g(x)\), $$ g''(x)=-\frac{-2}{(-1-2t)^3} =\frac{2}{-(1+2t)^3} =-\frac{2}{(1+2t)^3}<0. $$ The inverse is therefore also concave down. To check the formula directly, differentiate the explicit inverse: $$ g'(x)=-(1-4x)^{-1/2},\qquad g''(x)=-2(1-4x)^{-3/2}. $$ Because \(1+2g(x)=\sqrt{1-4x}\), this is exactly \(-2/(1+2t)^3\) when \(t=g(x)\). For a pointwise check, \(t=1\) maps to \(x=-2\). The formula gives \(-2/(1+2)^3=-2/27\); direct differentiation gives \(-2(1+8)^{-3/2}=-2/27\). The inverse and original function have the same concavity, as the decreasing-function sign rule predicts.
A Common Mistake: Ignoring Whether the Function Decreases
A tempting shortcut is to claim that the inverse always has the opposite concavity, because the formula begins with a minus sign. That overlooks the sign of the denominator. For example, with \(f(t)=e^{-t}\), one might incorrectly reason that \(f''(t)>0\) implies \(g''(x)<0\). But \(f'(t)<0\), so \((f'(t))^3<0\) as well. The correct calculation is $$ g''(x)=-\frac{x}{(-x)^3}=\frac{1}{x^2}>0, $$ not a negative value. The inverse is concave up, not concave down.
A reliable procedure is to determine both signs, \(f''\) and \(f'\), before deciding the sign of \(g''\). In particular, do not replace the denominator by \((f')^2\): that would lose the sign information that distinguishes increasing from decreasing functions.
Why the Relationship Is Useful
The formula lets you analyze inverse concavity without explicitly solving \(x=f(t)\) for \(t\). Finding an inverse formula can be difficult or impossible in elementary functions, but the original derivatives may still be easy to compute. The formula then identifies the inverse’s curvature at \(x=f(t)\) directly from \(f'(t)\) and \(f''(t)\).
It also makes the geometric effect of inversion precise. Reflection across \(y=x\) exchanges the roles of input and output, and the curvature relationship depends on the direction of travel along the original graph. The first derivative records that direction; the second derivative records the original concavity. Both are needed to determine the inverse’s concavity.
Check Your Understanding
For each question, state the relevant derivative signs and connect points in the domain of \(f\) with their images in the domain of \(f^{-1}\).
- State the conditions used to apply the second-derivative formula for an inverse.
- If \(f'(t)>0\) and \(f''(t)<0\) on an interval, what is the sign of the second derivative of the inverse at \(f(t)\)?
- If \(f'(t)<0\) and \(f''(t)<0\), is the inverse concave up or concave down at the corresponding point?
- For \(f(t)=1/t\) on \(t>0\), use the formula to find the second derivative of its inverse.
- Why does the minus sign in the inverse formula not imply that inverse concavity is always opposite to original concavity?
