Differentials — Question 1

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Question 1

Let f(x)=x3f(x) = x^3.

(a) Use differentials to approximate the change in ff as xx changes from 22 to 2.052.05.

(b) Compare your result to the actual change f(2.05)−f(2)f(2.05) - f(2).

Original worksheet page 1: question and worked solution for 4-12-001
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Question 1 - Solution

We are given: f(x)=x3,f′(x)=3x2f(x) = x^3, \quad f'(x) = 3x^2

(a) Use: df=f′(x)dxdf = f'(x) \, dx

Let x=2x = 2, dx=0.05dx = 0.05: f′(2)=3(2)2=12f'(2) = 3(2)^2 = 12

df=12⋅0.05=0.6df = 12 \cdot 0.05 = \boxed{0.6}

So, the approximate change in ff is 0.6\boxed{0.6}

(b) Actual change: f(2.05)=(2.05)3=8.615125,f(2)=8f(2.05) = (2.05)^3 = 8.615125, \quad f(2) = 8 Actual change=8.615125−8=0.615125\text{Actual change} = 8.615125 - 8 = \boxed{0.615125}

Conclusion: The differential estimate 0.60.6 is close to the actual change 0.6151250.615125

Original worksheet page 2: question and worked solution for 4-12-001

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