Answer
$$-0.0245,\ \ \ 0.00547,\ \ 22.30\%$$
Work Step by Step
Given $$f(x)=\frac{1}{1+x^{2}}, \quad a=3, \quad \Delta x=0.5$$ Since $$ f'(x) =\frac{-2x}{\left(1+x^2\right)^2} ,\ \ \ f'(3)=-0.06 $$ Then \begin{align*} \Delta f &\approx f^{\prime}(a) \Delta x\\ &=(0.5)(-0.06)\\ &= -0.03 \end{align*} Since the change $\Delta f$ is given by \begin{align*} \Delta f&=f(a+\Delta x)-f(a)\\ &=f(3.5)-f(3)\\ & \approx -0.0245 \end{align*} to find error $$|-0.03+0.0245|=0.00547$$ and the percentage $$\frac{0.00547}{0.0245} \times 100 \% \approx22.30\%$$
