Question

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

2. \(f\left( x \right) = {e^{3x}}\), \(a = 0\)

Short answer

Linear approximation is \(L\left( x \right) = 3x + 1\).

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 \(f\left( x \right) = {e^{3x}}\) and the point is \(a = 0\).

Now \(f'\left( x \right) = 3{e^{3x}}\). Hence slope of the curve at \(a = 0\) is \(f'\left( 0 \right) = 3\).

Now the equation of the tangent line at \(\left( {0,f\left( 0 \right)} \right)\) is,

\(\begin{aligned}y &= f\left( 0 \right) + f'\left( 0 \right)\left( {x - 0} \right)\\ &= {e^0} + 3\left( x \right)\\ &= 1 + 3x\end{aligned}\)

Hence the linear approximation is \(L\left( x \right) = 3x + 1\).