Answer
$1.80397$
Work Step by Step
The local quadratic approximation can be calculated as:
$f(x)=f(x_0) +f^{\prime}(x_0)(x-x_0)+f^{\prime\prime}(x_0)(x-x_0)^2$
Now, the local quadratic approximation of $f$ at $x=1$ is:
$ \tan 61^{\circ} \approx f(1) +f^{\prime}(1)(x-1)+f^{\prime\prime}(1)(x-1)^2\\ \approx 1.557408+3.426(1.064465-1)+10.66986(1.06465-1)^2\\ \approx 1.80397$
By using a calculator, we have: $ \tan 61^{\circ} \approx 1.804048$
Now, the absolute error is: $E=|1.804048-1.80397|=7.776 \times 10^{-5}$
