Use the local linear approximation of f(x) = the cube root of (64 + x) at x = 0 to approximate the cube root of 62.5, and compare the estimate with the exact value.

Step-by-step solution

1. 1

   Understand why the function is written as $\sqrt[3]{64+x}$ rather than $\sqrt[3]{x}$. Linear approximation is only accurate near the base point, and it is only useful when the base point is somewhere you can evaluate exactly. Here $64$ is a perfect cube, and $62.5$ is close to it, so shifting the variable puts the base point at $x_0=0$ with $f(0)=\sqrt[3]{64}=4$ known exactly.

2. 2

   Differentiate and evaluate the derivative at the base point. Writing $f(x)=(64+x)^{1/3}$,

   $f'(x)=\frac13(64+x)^{-2/3}=\frac{1}{3\sqrt[3]{(64+x)^2}}$

   $f'(0)=\frac{1}{3\cdot 64^{2/3}}=\frac{1}{3\cdot 16}=\frac{1}{48}$

3. 3

   Write down the linearisation. The tangent line at $x_0=0$ is

   $L(x)=f(0)+f'(0)(x-0)=4+\frac{x}{48}$

   This says: a small change $x$ in the radicand changes the cube root by roughly $x/48$. The factor $1/48$ is the whole content of the approximation.

4. 4

   Choose the input that produces the requested number. We need $64+x=62.5$, so

   $x=-1.5$

   $\sqrt[3]{62.5}\approx L(-1.5)=4+\frac{-1.5}{48}=4-0.03125=3.96875=\frac{127}{32}$

5. 5

   Compare with the exact value.

   $\sqrt[3]{62.5}=3.968503\ldots$

   $\text{absolute error}=3.96875-3.968503=0.000247,\qquad\text{relative error}\approx 0.0062\%$

   Six significant figures of agreement in the first four digits from one derivative evaluation.

6. 6

   Explain the direction of the error, which is not an accident. Differentiating again,

   $f''(x)=-\frac{2}{9}(64+x)^{-5/3}<0$

   so $f$ is concave down near $x=0$ and its graph lies below every tangent line. The linear estimate must therefore be an overestimate — and indeed $3.96875>3.968503$. The same reasoning predicts the error grows like $\tfrac12|f''(0)|x^2$, roughly $0.00024$ for $x=-1.5$, matching the observed error almost exactly.