Reflect the Point First, Then Take the Reciprocal

You'll learnto read an inverse slope straight off the mirror image, and to see why feeding the outer derivative the wrong number draws a line that is tangent to nothing.

The graph of an inverse is the original reflected across the line y = x, and reflecting a line exchanges its rise with its run. A tangent of slope 3 comes back as a tangent of slope 1/3. That is the whole theorem, and it is visible: draw the triangle, slide it across the mirror, and watch the legs swap. What is genuinely hard is not the reciprocal but the address — which point the reciprocal is taken at. The slope of the inverse at x = 5 is the reciprocal of the slope of f at x = 1, because that is the point 5 reflects onto. Take the reciprocal at the wrong point and you get a plausible number attached to nothing.

8 STEPS · 6 QUICK CHECKS · (f⁻¹)′(b) = 1 / f′(a) WHERE f(a) = b · THE ADDRESS, NOT THE ARITHMETIC · v1

f(a) = b ⇄ (f⁻¹)′(b) = 1 / f′(a)

Before you start

What you're looking at

A function in blue, its inverse in pink, and the dashed line y = x between them. The two curves are mirror images: every point on one has a twin on the other with the coordinates swapped.

The question

If f climbs 3 units for every 1 across at a point, what does its mirror image do at the mirrored point — and at which point exactly?

Watch for

Step 2, where a slope triangle slides across the mirror and its legs trade places — and step 7, where the wrong input draws a line that touches nothing.

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