Question

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

1. \(f\left( x \right) = {x^3} - {x^2} + 3\), \(a = - 2\)

Short answer

Linear approximation is \(L\left( x \right) = 16x + 23\).

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 \(f\left( x \right) = {x^3} - {x^2} + 3\) and the point is \(a = - 2\).

Now \(f'\left( x \right) = 3{x^2} - 2x\). Hence slope of the curve at \(a = - 2\) is \(f'\left( { - 2} \right) = 16\).

Now the equation of the tangent line at \(\left( { - 2,f\left( { - 2} \right)} \right)\) is,

\(\begin{aligned}y &= f\left( { - 2} \right) + f'\left( { - 2} \right)\left( {x + 2} \right)\\ &= \left( {{{\left( { - 2} \right)}^3} - {{\left( { - 2} \right)}^2} + 3} \right) + 16\left( {x + 2} \right)\\ &= \left( { - 8 - 4 + 3} \right) + \left( {16x + 32} \right)\\ &= - 9 + 16x + 32\\&= 16x + 23\end{aligned}\)

Hence the linear approximation is \(L\left( x \right) = 16x + 23\).