Why Test Before Using the Formula?
An inverse derivative is related to the slope of the original function: when the original function has a nonzero slope, the inverse’s slope at the corresponding point is its reciprocal. But the formula is not a substitute for checking that the inverse value exists or that the original slope meets the required condition.
For a specified value \(y_0\), the important checks are connected. First, \(y_0\) must be in the domain of the inverse, which is the range of the original function on its selected domain. Next, find the original input \(x_0\) for which \(f(x_0)=y_0\). Then check \(f'(x_0)\). The reciprocal-slope formula applies when this derivative exists and is nonzero, with the inverse defined on the chosen branch.
The Inverse-Derivative Test
A function must be one-to-one on its selected domain to have an inverse function there. If the original function is not one-to-one on its full domain, a restricted domain, or branch, may give a valid inverse. The chosen branch matters: it determines which original input corresponds to a given inverse input.
These are sufficient conditions for using the formula. In particular, \(y_0\) must belong to \(f(I)\), the domain of the inverse. The condition \(f'(x_0)\ne 0\) is essential for this reciprocal formula: it prevents division by zero and ensures that the resulting slope is finite.
The test is point-specific. A function may have an inverse and an inverse derivative at some values but not others. Even if the inverse exists throughout its domain, the reciprocal formula must be checked at the particular corresponding input.
Why the Reciprocal-Slope Formula Works
Let \(g=f^{-1}\). At corresponding points, \(g(y_0)=x_0\) and \(f(x_0)=y_0\). For \(y\ne y_0\) in the inverse’s domain, set \(x=g(y)\). Then \(f(x)=y\), so the difference quotient for the inverse can be rewritten as a reciprocal:
As \(y\) approaches \(y_0\), continuity of the inverse gives \(x=g(y)\to x_0\). Because \(f\) is differentiable at \(x_0\), the denominator in the final expression approaches \(f'(x_0)\). If that limit is nonzero, taking the reciprocal is valid, and the limit of the inverse difference quotient is \(1/f'(x_0)\). This proves the formula under the theorem’s conditions.
Worked Example: A Quadratic on a Restricted Domain
Worked Example: A Quadratic on a Restricted Domain
Let \(f(x)=x^2-4x+5\) with domain \(I=[2,\infty)\). Find \((f^{-1})'(5)\), or explain why the reciprocal-slope formula cannot be used.
Step 1: Check the inverse’s domain. Complete the square:
Since \(x\ge 2\), \((x-2)^2\ge 0\), so \(f(x)\ge 1\). The range of \(f\), and thus the domain of \(f^{-1}\), is \([1,\infty)\). Since \(5\ge 1\), the requested inverse input is in its domain.
Step 2: Find the corresponding original input.
The equation gives \(x=0\) or \(x=4\), but the selected domain requires \(x\ge 2\). Thus the corresponding input is \(x_0=4\), and \(f^{-1}(5)=4\).
Step 3: Check the original derivative and apply the formula.
The derivative exists and is nonzero, so the formula applies:
Check: On the selected domain, \(f^{-1}(y)=2+\sqrt{y-1}\). Therefore
This agrees with the reciprocal-slope calculation.
When a Check Fails
Each check has a distinct purpose. If the requested value is outside the inverse’s domain, there is no corresponding inverse value to differentiate. If the original input is not uniquely determined on the selected domain, there is no inverse function there. If the original derivative at the corresponding input is zero, the reciprocal formula is undefined; that does not permit substituting zero into the denominator or claiming a finite inverse derivative.
Worked Example: The Original Derivative Is Zero
Let \(f(x)=x^3\) on \(\mathbb{R}\). Test whether the reciprocal-slope formula gives \((f^{-1})'(0)\).
Step 1: Check the inverse’s domain. The function \(x^3\) takes every real value and is one-to-one on \(\mathbb{R}\). Therefore its inverse is defined at \(0\).
Step 2: Find the corresponding input.
Step 3: Check the original derivative.
The derivative is zero, so the reciprocal-slope formula does not apply: it would require division by zero. To see what happens to the inverse itself, \(f^{-1}(y)=\sqrt[3]{y}\). Its difference quotient at zero is
As \(y\to 0\), this quotient grows without bound, so the inverse has no finite derivative at \(0\). Here the failed nonzero-derivative check correctly warns that the reciprocal formula cannot produce a finite slope.
Worked Example: The Requested Value Is Outside the Inverse’s Domain
Let \(f(x)=\sqrt{x-1}\) with domain \(x\ge 1\). Can the reciprocal-slope formula be used to find \((f^{-1})'(-2)\)?
Step 1: Determine the range of \(f\). A square root is nonnegative, so \(f(x)\ge 0\). As \(x\) ranges over \([1,\infty)\), \(f(x)\) ranges over \([0,\infty)\). This is the domain of \(f^{-1}\).
Because \(-2\notin[0,\infty)\), \(f^{-1}(-2)\) is not defined. Equivalently, solving for a corresponding original input would require
which has no solution in the original domain. Therefore \((f^{-1})'(-2)\) does not exist, and there is no corresponding input at which to evaluate \(f'\). The test stops at the domain check; a reciprocal-slope calculation is not appropriate.
A Common Mistake: Ignoring the Selected Branch
Consider \(f(x)=x^2\) on all of \(\mathbb{R}\). A tempting calculation for an inverse derivative at \(y=9\) is to select \(x=3\), compute \(f'(3)=6\), and report \(1/6\). But this original function is not one-to-one on \(\mathbb{R}\): both \(3\) and \(-3\) map to \(9\). There is no inverse function on the full domain, so the calculation has not established an inverse derivative.
Restricting the domain resolves the issue, but the choice must be stated. On \([0,\infty)\), the inverse is \(f^{-1}(y)=\sqrt{y}\), and the corresponding input is \(3\). The derivative is \(f'(3)=6\), so the inverse derivative at \(9\) is \(1/6\). On \((-\infty,0]\), the inverse is \(f^{-1}(y)=-\sqrt{y}\), and the corresponding input is \(-3\). Now \(f'(-3)=-6\), so the inverse derivative at \(9\) is \(-1/6\). The branch changes the answer.
A Reliable Order for Testing
Use the checks in order. This prevents wasted calculations and helps you distinguish an undefined inverse value from a failure of the derivative condition.
Identify the interval or branch on which the original function is one-to-one and the inverse is being defined.
Determine the range of the original function on that domain. If \(y_0\) is outside the range, stop: \(f^{-1}(y_0)\) is undefined.
Solve \(f(x_0)=y_0\) and select the solution in the chosen domain. This gives \(x_0=f^{-1}(y_0)\).
Compute \(f'(x_0)\), not \(f'(y_0)\). Check that the derivative exists and is nonzero.
If all conditions hold, calculate \((f^{-1})'(y_0)=1/f'(x_0)\), then verify that the corresponding inputs and values have been used.
Why the Test Matters
The inverse-derivative formula appears simple, but its inputs can be easy to confuse. The original derivative is evaluated at \(x_0\), the input that maps to the requested inverse input \(y_0\). The inverse derivative is then evaluated at \(y_0\). Keeping these roles separate prevents the common error of differentiating the original function at the wrong value.
The test also clarifies what a failed calculation means. An out-of-range value has no inverse value. A missing branch choice can make the inverse ambiguous. A zero original derivative makes the reciprocal formula unusable, and the inverse may fail to have a finite derivative there. These are different situations, so state which condition fails rather than treating every difficulty as the same kind of error.
Check Your Understanding
For each question, identify the relevant domain, input, or derivative check before deciding whether the reciprocal-slope formula applies.
- For \(f(x)=x^2\) restricted to \([0,\infty)\), is \((f^{-1})'(4)\) defined by the reciprocal-slope formula? Identify the corresponding original input.
- For \(f(x)=\sqrt{x+2}\) on \(x\ge -2\), is \(f^{-1}(-1)\) defined?
- For \(f(x)=x^3+1\) on \(\mathbb{R}\), find the corresponding original input for inverse input \(1\), and check whether the reciprocal-slope formula applies.
- Why must a branch be specified before using \(f(x)=x^2\) to find an inverse derivative?
- If \(f'(x_0)=0\), what does that tell you about using the reciprocal-slope formula, and what does it not automatically tell you about every possible inverse situation?
