Question

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

4. \(f\left( x \right) = cos2x\), \(a = \frac{\pi }{6}\)

Short answer

Linear approximation is \(L\left( x \right) = - \sqrt 3 x + \frac{{\sqrt 3 \pi }}{6} + \frac{1}{2}\).

Step-by-step solution

Linear Approximation of a Curve at a Given point

We can approximate a 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 \(f\left( x \right) = \cos 2x\) and the point is \(a = \frac{\pi }{6}\).

Now \(f'\left( x \right) = - 2\sin 2x\). Hence slope of the curve at \(a = \frac{\pi }{6}\) is, \(f'\left( {\frac{\pi }{6}} \right) = - 2\sin \frac{\pi }{3} = - \sqrt 3 \).

Now the equation of the tangent line at \(\left( {\frac{\pi }{6},f\left( {\frac{\pi }{6}} \right)} \right)\) is,

\(\begin{aligned}y &= f\left( {\frac{\pi }{6}} \right) + f'\left( {\frac{\pi }{6}} \right)\left( {x - \frac{\pi }{6}} \right)\\ &= \frac{1}{2} + \left( { - \sqrt 3 } \right)\left( {x - \frac{\pi }{6}} \right)\\ &= \frac{1}{2} - \sqrt 3 x + \frac{{\sqrt 3 \pi }}{6}\end{aligned}\)

Hence the linear approximation is \(L\left( x \right) = - \sqrt 3 x + \frac{{\sqrt 3 \pi }}{6} + \frac{1}{2}\).