If $f(x) = p \sin x + qx \cos x + x^2$ is defined for all $x,q,p \in \Bbb R$ and $f(2) = 3$ find $f(-2)$.

Hint:

$$f(x)+f(-x)=(p\sin x + qx \cos x + x^2) + (p\sin(-x) + q(-x)\cos (-x) + (-x)^2\\=p\sin x + qx\cos x + x^2 - p\sin x - qx \cos x + x^2$$

Now simplify the equation (a lot of things cancel out) and plug in $x=2$.

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An alternative would be to directly write

$$f(-2)=p\sin(-2) + -2q\cos 2 + 4 = (-p\sin(2) - 2q\cos 2 - 2^2) + 2^2 + 4 = -f(2)+8$$

$$f\left( -2 \right) =p\sin { \left( -2 \right) -2q\cos { 2 } +4 } =-p\sin { 2 } -2q\cos { 2 } +4=\\ =-\left( \underset { -1 }{ \underbrace { p\sin { 2+ } 2q\cos { 2 } } } \right) +4=\color{red}{5}$$

We are given that $f(x)=u(x)+x^2$ with $u$ an odd function. From $3=f(2)=u(2)+2^2$ it follows that $u(2)=-1$, hence $f(-2)=-u(2)+(-2)^2=5$.