The slope of the function

2017

The slope of the function

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Answer: B. 0ConceptThe slope of a curve at a point is the value of its derivative there. When a function is defined piecewise and the point lies at the join, the…

  1. A.

    1

  2. B.

    0

  3. C.

    -1

  4. D.

    None

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Correct answer: B

Concept

The slope of a curve at a point is the value of its derivative there. When a function is defined piecewise and the point lies at the join, the derivative cannot be read off a standard rule; it must be computed from the limit definition f'(a) = limh→0 [ f(a+h) − f(a) ] / h. A bounded factor times a quantity tending to 0 also tends to 0 (Squeeze Theorem): since −1 ≤ sin θ ≤ 1, |h·sin θ| ≤ |h| → 0.

Application

We need the slope at x = 0, i.e. f'(0), with f(0) = 0.

  1. Write the difference quotient at 0: [f(0+h) − f(0)] / h = [ h2 sin(1/h) − 0 ] / h.

  2. Cancel one factor of h: = h · sin(1/h).

  3. Bound it: since |sin(1/h)| ≤ 1, we have 0 ≤ |h·sin(1/h)| ≤ |h|.

  4. Take the limit as h → 0: |h| → 0, so by the Squeeze Theorem h·sin(1/h) → 0. Hence f'(0) = 0.

The slope of the function at the origin is therefore 0.

Cross-check

For x ≠ 0 the product rule gives f'(x) = 2x·sin(1/x) − cos(1/x). This does not approach a single limit as x → 0 (the cos(1/x) term oscillates), so the derivative is NOT continuous at 0 — yet the value at 0 itself, obtained from the definition above, exists and equals 0. The two facts are consistent: a derivative can exist at a point without being continuous there.

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