Question

1-4: Find the linearization \(L\left( x \right)\) of the function at \(a\).

3. \(f\left( x \right) = \sqrt[3]{x}\), \(a = 8\)

Short answer

Linear approximation is \(L\left( x \right) = \frac{1}{{12}}x + \frac{4}{3}\).

Step-by-step solution

Linear Approximation of a Curve at a Given point

We can approximatea function \(f\left( x \right)\) of a curve at a point \(a\) by the tangent line at the given point \(\left( {a,f\left( a \right)} \right)\). To do that first we have to find the equation of the tangent line at that given point, which is,

\(y - f\left( a \right) = f'\left( a \right)\left( {x - a} \right)\)

Hence the linear approximation of the curve is, \(L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right)\).

Finding the Linear approximation of the Curve

Given that \(f\left( x \right) = \sqrt[3]{x}\) and the point is \(a = 8\).

Now \(f'\left( x \right) = \frac{1}{3}{x^{\left( {\frac{{ - 2}}{3}} \right)}}\). Hence slope of the curve at \(a = 8\) is,

\(\begin{aligned}f'\left( { - 2} \right) &= \frac{1}{3}\frac{1}{{\sqrt[3]{{{8^2}}}}}\\ &= \frac{1}{3} \times \frac{1}{4}\\ &= \frac{1}{{12}}\end{aligned}\).

Now the equation of the tangent line at \(\left( {8,f\left( 8 \right)} \right)\) is,

\(\begin{aligned}y &= f\left( 8 \right) + f'\left( 8 \right)\left( {x - 8} \right)\\ &= \sqrt[3]{8} + \frac{1}{{12}}\left( {x - 8} \right)\\ &= 2 + \frac{1}{{12}}x - \frac{8}{{12}}\\ &= \frac{1}{{12}}x + 2 - \frac{2}{3}\\ &= \frac{1}{{12}}x + \frac{4}{3}\end{aligned}\)

Hence the linear approximation is \(L\left( x \right) = \frac{1}{{12}}x + \frac{4}{3}\).