The linear approximation to f ( x ) = x 3 f(x)=\sqrt[3]{x} f ( x ) = 3 x at x 0 = 9 x_{0}=9 x 0 = 9 .
A linear approximation is given by L ( x ) ≈ f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) . L(x) \approx f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right) . L ( x ) ≈ f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) .
We are given that x 0 = 9 x_{0}=9 x 0 = 9 .
Firstly, find the value of the function at the given point: y 0 = f ( x 0 ) = 3 2 3 y_{0}=f\left(x_{0}\right)=3^{\frac{2}{3}} y 0 = f ( x 0 ) = 3 3 2 .
Secondly, find the derivative of the function, evaluated at the point: f ′ ( 9 ) f^{\prime}(9) f ′ ( 9 ) .
Find the derivative: f ′ ( x ) = 1 3 x 2 3 f^{\prime}(x)=\frac{1}{3 x^{\frac{2}{3}}} f ′ ( x ) = 3 x 3 2 1
Next, evaluate the derivative at the given point to find slope.
f ′ ( 9 ) = 3 2 3 27 f^{\prime}(9)=\frac{3^{\frac{2}{3}}}{27} f ′ ( 9 ) = 27 3 3 2
Plugging the values found, we get that L ( x ) ≈ 3 2 3 + 3 2 3 27 ( x − ( 9 ) ) L(x) \approx 3^{\frac{2}{3}}+\frac{3^{\frac{2}{3}}}{27}(x-(9)) L ( x ) ≈ 3 3 2 + 27 3 3 2 ( x − ( 9 )) .
Or, more simply: L ( x ) ≈ 3 2 3 27 x + 2 ⋅ 3 2 3 3 L(x) \approx \frac{3^{\frac{2}{3}}}{27} x+\frac{2 \cdot 3^{\frac{2}{3}}}{3} L ( x ) ≈ 27 3 3 2 x + 3 2 ⋅ 3 3 2 .
Answer: L ( x ) ≈ 3 2 3 27 x + 2 ⋅ 3 2 3 3 ≈ 0.077040141594515 x + 1.38672254870127 L(x) \approx \frac{3^{\frac{2}{3}}}{27} x+\frac{2 \cdot 3^{\frac{2}{3}}}{3} \approx 0.077040141594515 x+1.38672254870127 L ( x ) ≈ 27 3 3 2 x + 3 2 ⋅ 3 3 2 ≈ 0.077040141594515 x + 1.38672254870127 .
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