The linearization is .
The approximation is .
In this lesson we will use the tangent line to approximate the value of a function near the point of tangency.
Given a function , the equation of the tangent line at the point where is given by or The main idea of this section is that if we let then and for values of close to . The function is called the linearization of at . The advantage of working with is that values of a linear function are usually easy to compute. In a typical linear approximation problem, we are trying to approximate a value of . We need to choose and create . Once we have accomplished this, our solution is There are two keys to choosing . First should be close to the -value of interest and second, we must be able to compute the exact value of .
The linearization of at is now given by \begin{align*} L(x) &= f(a) + f'(a)(x - a) \\ &= f(4) + f'(4)(x - 4) \\ &= 2 + \frac{1}{4}(x - 4). \end{align*}
Finally, if is near 4, then , so
The linearization of at is now given by \begin{align*} L(x) &= f(a) + f'(a)(x - a) \\ &= f(100) + f'(100)(x - 100) \\ &= 10 + \frac{1}{20}(x - 100). \end{align*}
Finally, if is near 100, then , so
The linearization is .
The approximation is .
The linearization is .
The approximation is .
Next, \begin{align*} L(x) &= f(a) + f'(a)(x - a) \\ &= f(8) + f'(8)(x - 8) \\ &= 2 + \frac{1}{12}(x - 8) \end{align*}
and finally,
The linearization is .
The approximation is .
Next, \begin{align*} L(x) &= f(a) + f'(a)(x - a) \\ &= f(0) + f'(0)(x - 0) \\ &= 1 + 1(x - 0)\\ & = 1 + x \end{align*}
and finally,
We conclude this example by computing relative error. Expressed as a percentage, the relative error is given by For the actual value, we will use the value of from a calculator. We have
The linearization is .
The approximation is .
The relative error in this approximation is (to two decimal places) .
Next, \begin{align*} L(x) &= f(a) + f'(a)(x - a) \\ &= f(1) + f'(1)(x - 1) \\ &= 0 + 1(x - 1)\\ &=x-1 \end{align*}
and finally,
The linearization is .
The approximation is .
Next, \begin{align*} L(x) &= f(a) + f'(a)(x - a) \\ &= f(2) + f'(2)(x - 2) \\ &= 8 + 12(x - 2) \end{align*}
and finally,
The linearization is .
The approximation is .
We let and since 0 is near and , we let . To create we also need to compute . Since we have
Next, \begin{align*} L(x) &= f(a) + f'(a)(x - a) \\ &= f(0) + f'(0)(x - 0) \\ &= 0 + 1(x - 0)\\ &= x \end{align*}
and finally,