Linear approximation practice question 6

The prompt hands you $\arctan 1$. Which base point does that suggest, and is 0.9 to its left or to its right?

Base point $a = 1$: $\dfrac{d}{dx}\arctan x = \dfrac{1}{1 + x^{2}}$ equals $\tfrac{1}{2}$ there.

$\arctan(0.9) \approx \arctan 1 + \tfrac{1}{2}(0.9 - 1)$.