Changing Rate Math Example 2

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Example 2

hard
For g(x)=x3g(x) = x^3, find the average rate of change on [a,a+h][a, a+h] and simplify to see what happens as h→0h \to 0.

Solution

  1. 1
    Compute: g(a+h)−g(a)h=(a+h)3−a3h\frac{g(a+h)-g(a)}{h} = \frac{(a+h)^3 - a^3}{h}.
  2. 2
    Expand: (a+h)3=a3+3a2h+3ah2+h3(a+h)^3 = a^3 + 3a^2h + 3ah^2 + h^3. Subtract and divide: 3a2h+3ah2+h3h=3a2+3ah+h2\frac{3a^2h+3ah^2+h^3}{h} = 3a^2 + 3ah + h^2.
  3. 3
    As h→0h \to 0: 3a2+3a(0)+02=3a23a^2 + 3a(0) + 0^2 = 3a^2. This is the derivative g′(a)=3a2g'(a) = 3a^2, the instantaneous rate of change.

Answer

Average rate =3a2+3ah+h2= 3a^2+3ah+h^2; instantaneous rate g′(a)=3a2g'(a)=3a^2
The average rate of change over [a,a+h][a, a+h] is the difference quotient. Taking the limit h→0h\to0 gives the derivative, connecting average rate (secant slope) to instantaneous rate (tangent slope).

About Changing Rate

A changing rate of change means the output grows by different amounts for equal increases in input — the hallmark of nonlinear functions like quadratics and exponentials.

Learn more about Changing Rate →

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Example 4 medium

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