Math CentralQuandaries & Queries


Question from Amit, a teacher:

consider the equation = x^3. The equation of tangent to this curve (which is symmetrical in Ist and IVth quadrant) at (0,0) is y=0, which is x-axis.
but graphically one can visualize that x-axis intersects the curve, so how can it be the tangent to the curve. Please help.

Hi Amit,

The challenge here is to make a precise definition of the tangent line to a curve at a point. This is one of the first major steps in learning calculus. The difficulty arises because the tangent line can't be defined using only algebraic concepts. The procedure is as follows.

Take a curve y = f(x) (in your case y = f(x) = x3) and a point P on the graph of the function (you have P = (0,0)). Let Q be some other point on the curve and draw the line through P and Q. This line is called a secant line PQ. Now keep P where it is but let Q move along the curve towards P and watch the secant line PQ as Q moves.

As Q approaches P the secant line PQ approaches what I think should be the tangent line, the tangent line arises when P = Q. But this doesn't make sense since when P = Q "the line through P and Q" is ambiguous. This is the first major challenge of calculus. How do you make mathematically precise the ides that the tangent line is the line that the secant line PQ approaches as Q approaches P?

Once you do this you find that the tangent line to f(x) = x3 at P = (0,0) is the line y = 0.

I hope this helps,

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