Answers f ′ ( x ) = 1 1 + ( 2 + x ) 2 f'(x) = \frac{1}{1 + (\sqrt{2} + x)^2} f ′ ( x ) = 1 + ( 2 + x ) 2 1
f ′ ′ ( x ) = − 2 ( 2 + x ) ( 1 + ( 2 + x ) 2 ) 2 f''(x) = \frac{-2(\sqrt{2} + x)}{\left( 1 + (\sqrt{2} + x)^2 \right)^2} f ′′ ( x ) = ( 1 + ( 2 + x ) 2 ) 2 − 2 ( 2 + x )
0.955 + 0.333 x − 0.157 x 2 + … 0.955 + 0.333x -0.157x^2 + \ldots 0.955 + 0.333 x − 0.157 x 2 + … Full solutions
(a) f ′ ( x ) = 1 1 + ( 2 + x ) 2 ■ f ′ ′ ( x ) = − 2 ( 2 + x ) ( 1 + ( 2 + x ) 2 ) 2 ■ \begin{align*}
f'(x) &= \frac{1}{1 + (\sqrt{2} + x)^2} \; \blacksquare
\\ f''(x) &= \frac{-2(\sqrt{2} + x)}{\left( 1 + (\sqrt{2} + x)^2 \right)^2} \; \blacksquare
\end{align*} f ′ ( x ) f ′′ ( x ) = 1 + ( 2 + x ) 2 1 ■ = ( 1 + ( 2 + x ) 2 ) 2 − 2 ( 2 + x ) ■
(b) f ( 0 ) = tan − 1 2 = 0.955 (3 sf) \begin{align*}
f(0) &= \tan^{-1} \sqrt{2}
\\ &= 0.955 \textrm{ (3 sf)}
\end{align*} f ( 0 ) = tan − 1 2 = 0.955 (3 sf) f ′ ( 0 ) = 1 1 + 2 = 1 3 = 0.333 (3 sf) \begin{align*}
f'(0) &= \frac{1}{1+2}
\\ &= \frac{1}{3}
\\ &= 0.333 \textrm{ (3 sf)}
\end{align*} f ′ ( 0 ) = 1 + 2 1 = 3 1 = 0.333 (3 sf) f ′ ′ ( 0 ) = − 2 2 ( 1 + 2 ) 2 = − 0.31427 (5 sf) \begin{align*}
f''(0) &= \frac{-2\sqrt{2}}{(1+2)^2}
\\ &= -0.31427 \textrm{ (5 sf)}
\end{align*} f ′′ ( 0 ) = ( 1 + 2 ) 2 − 2 2 = − 0.31427 (5 sf)
Maclaurin series for
f ( x ) , {f(x),} f ( x ) , f ( x ) = 0.955 + 0.333 x + − 0.31427 2 x 2 + … = 0.955 + 0.333 x − 0.157 x 2 + … ■ (3 sf) \begin{align*}
f(x) &= 0.955 + 0.333x + \frac{-0.31427}{2} x^2 + \ldots
\\ &= 0.955 + 0.333x -0.157x^2 + \ldots \; \blacksquare \textrm{ (3 sf)}
\end{align*} f ( x ) = 0.955 + 0.333 x + 2 − 0.31427 x 2 + … = 0.955 + 0.333 x − 0.157 x 2 + … ■ (3 sf) Question Commentary
As this is a calculus question, we need to ensure that we are working in radians rather than degrees: this
is likely the source of most mistakes.
Otherwise, this is a reasonably standard application of differentiation techniques
(including the tangent inverse differentiation formula that is provided in the formula booklet)
and the general Maclaurin formula (also provided) involving derivatives.