Answer:
[tex](\pi - 8)[/tex].
Step-by-step explanation:
In general, for constant [tex]a[/tex] (for which both [tex]f(a)[/tex] and [tex]f^{\prime}(a)[/tex] are known,) the linear approximation of [tex]f(x)[/tex] would be:
[tex]f(a) + (x - a)\, f^{\prime}(a)[/tex].
In this question, the constant is [tex](\pi / 2)[/tex].
Substituting this value into the expression gives [tex]f(\pi / 2) = 2\, \cos( \pi / 2) + 1 = 1[/tex].
Differentiate [tex]f[/tex] to obtain [tex]f^{\prime}(x) = (-2)\, \sin(x)[/tex]. Substitute in [tex]x = (\pi / 2)[/tex] to get [tex]f^{\prime}(\pi / 2) = (-2)\, \sin(\pi /2) = (-2)[/tex].
Thus, the linear approximation of [tex]f(3)[/tex] using [tex]f(\pi / 2)[/tex] and [tex]f^{\prime}(\pi / 2)[/tex] would be:
[tex]\begin{aligned} & f\left(\frac{\pi}{2}\right) + \left(3 - \frac{\pi}{2} \right)\, f^{\prime}\left(\frac{\pi}{2}\right) \\ =\; & 1 + \left(3 - \frac{\pi}{2}\right)\times (-2) \\ =\; & 1 + (-6) + \pi \\ =\; & \pi - 5\end{aligned}[/tex].
